Physics Labs
LJ413
hwm.zipLab 2/Lab 2- Kinematics.pdf
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Learning Objec뙕ves
Disᣊ�nguish between scalar and vector quanᣊ�ᣊ�es Apply kinemaᣊ�c equaᣊ�ons to 1‐D and projecᣊ�le moᣊ�on Predict posiᣊ�on, velocity, and acceleraᣊ�on vs. ᣊ�me graphs Calculate average and instantaneous velocity or acceleraᣊ�on Determine that x and y components are independent of each other Relate velocity, radius, and ᣊ�me period to uniform circular moᣊ�on. Explain the direcᣊ�on of acceleraᣊ�on during uniform circular moᣊ�on
1‐D Kinema뙕cs
1‐D kinemaᣊ�cs occurs when an object travels in one dimension and can be described using words, equaᣊ�ons and graphs. Linear mo뙕on describes how an object will move horizontally or verᣊ�cally with constant acceleraᣊ�on, how an object will
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Figure 1: Pool balls in moᾷon demonstrate 1‐D kinemaᾷcs.
Figure 2: Line secant to the path of the object.
travel if dropped from the side of a cliff, and the path it will follow if thrown straight up into the air. Keep in mind the moᣊ�on of an object is relaᣊ�ve to the viewer. Even though you do not feel like you are in moᣊ�on right now, you are on planet earth that has rotaᣊ�onal moᣊ�on in addiᣊ�on to orbital moᣊ�on around the sun. In almost all cases here moᣊ�on will be relaᣊ�ve to the Earth.
Scalar and Vector Quan뙕뙕es
In physics, quanᣊ�ᣊ�es can be scalar or vector. The difference between the two lies in direcᣊ�on. Scalar quanᣊ�ᣊ�es include magnitudes, which are numerical measurements. The distance an object has traveled or the speed of an object is a scalar quanᣊ�ty. Scalars do not take direcᣊ�on into consideraᣊ�on and can be described with only a number and a unit. For example, somebody might say the temperature outside is 70°F. Seventy is the magnitude, and Fahrenheit is the unit; there is no direcᣊ�on associated with the quanᣊ�ty. Vector quanᣊ�ᣊ�es, on the other hand, include magnitude and direcᣊ�on. The displacement from an object's iniᣊ�al posiᣊ�on, velocity, and acceleraᣊ�on are vector quanᣊ�ᣊ�es. The direcᣊ�on of vectors can be described as being in the posiᣊ�ve direcᣊ�on, in the negaᣊ�ve direcᣊ�on, north, south, east, west, leĀ, right, up, down, etc. One might describe an airplane's velocity as 450 miles per hour due west where both magnitude and direcᣊ�on are given. It is important to disᣊ�nguish between scalar and vector quanᣊ�ᣊ�es when trying to understand kinemaᣊ�cs.
Speed, Velocity, and Accelera뙕on
You may be familiar with speed outside of the physics classroom. When you drive in a car you are traveling a distance over a certain amount of ᣊ�me: a speed. How then is velocity different from speed? Velocity (v) is a vector quanᣊ�ty described as the rate at which an object's posiᣊ�on changes divided by the ᣊ�me the object is in moᣊ�on. Furthermore, the rate of change in velocity plays an important role in physics. Accelera뙕on represents the rate of change of an object's velocity over ᣊ�me.
Speed, Average Velocity, and Average Accelera뙕on
Before we define speed, velocity, and acceleraᣊ�on, it is important to disᣊ�nguish between distance and displacement. Distance is how much “ground” an object has covered; whereas, displacement is how far an object has moved from its original posiᣊ�on. The total change in an object's distance over ᣊ�me is referred to as average speed. Distance and speed are scalar quanᣊ�ᣊ�es. The average change in an object's displacement over ᣊ�me is referred to as average velocity, and the average change in an object's velocity over ᣊ�me is referred to as average acceleraᣊ�on. Displacement, velocity, and acceleraᣊ�on, as menᣊ�oned before, are vector quanᣊ�ᣊ�es. When you are calculaᣊ�ng the average velocity or acceleraᣊ�on of an object, you
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Figure 3: Diagram showing the average speed of a car. Speed is scalar.
Figure 4: Diagram showing the velocity of a car.
are actually finding the slope of the line secant to the path between the two points in ᣊ�me (Figure 2).
Example 1: Speed
A car travels a distance of 800 meters over a ᣊ�me of 10 seconds. The car has an average speed of 80 meters per second (Figure 3).
Speed AVG = distance
= 800 m
= 80 m
= 80 m/s 뙕me 10 s s
Example 2: Velocity
A car travels at 800 meters west. The car makes a U‐ turn and travels a distance of 800 meters east. It took the car 20 seconds to complete its course. The average velocity of the car is 0 meters per second (Figure 4) because the car travels in opposite direcᣊ�ons that cancel each other out.
VelocityAVG(vAVG) = Δdisplacement
= 800 m + (‐800) m
= 0 m
= 0 m/s 뙕me 20 s 20 s
Even though the car experienced a lot of moᣊ�on in 20 seconds, its posiᣊ�on did not change over the course of its moᣊ�on.
Example 3: Accelera뙕on
A car travels from 0 m/s to 90 m/s in 15 seconds. The car has an average acceleraᣊ�on of 6 m/s2.
Accelera뙕onAVG(aAVG) = Δvelocity
= 90 m/s
= 6 m/s
= 6 m/s2뙕me 15 s s
Example 4: Instantaneous Velocity
A car is moving along a path whose displacement can be modeled by funcᣊ�on x(t) = 2.0 m + (4.0 m/s2)t2 (Figure 5‐leĀ). The car's velocity is the derivaᣊ�ve of its displacement, v(t)= (8.0 m/s2)t (Figure 5‐right). The instantaneous velocity of the car at 3 seconds is 24 m/s.
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Figure 5: The car's displacement over a period of ᾷme is graphed to the le뎌. The car's velocity over a period if ᾷme is graphed to the right.
Displacement = x(t) = 2.0 m + (4.0 m/s2)t2
Velocity = (d/dt) x(t) = v(t) = (8.0 m/s2)t
v(3) = (8.0 m/s2) (3.0 s) = 240 m/s
Example 5: Instantaneous Accelera뙕on
A car is moving with a velocity that can be modeled by the funcᣊ�on v(t)= (8.0 m/s2)t. The car's acceleraᣊ�on is the derivaᣊ�ve of the velocity, a(t)= 8.0 m/s2 (Figure 6). The instantaneous acceleraᣊ�on of the car at any point in ᣊ�me is 8.0 m/s2 because in this case acceleraᣊ�on is not dependent on ᣊ�me. In other words, the acceleraᣊ�on is constant. For example, the acceleraᣊ�on at 5 seconds is 8.0 m/s2.
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Figure 6: Acceleraᾷon over ᾷme of the car.
Velocity = v(t) = 8.0 m/s2)t
Accelera뙕on = (d/dt) v(t) = (8.0 m/s2)
α(5) = (8.0 m/s2)
Graphing 1‐D Mo뙕on
In science, we observe how one factor changes as a result of a change in another. The effect of one variable on another can be expressed as a funcᣊ�on. An alternaᣊ�ve to using equaᣊ�ons to describe moᣊ�on is to uᣊ�lize moᣊ�on graphs to visualize the same relaᣊ�onships.
When we look at the speed vs. ᣊ�me graph of an object in free fall (Figure 7), we observe a linear relaᣊ�onship. In other words, for every increase in ᣊ�me, there is the same increase in speed. Since the object is dropped from rest, the line starts at the origin (0,0). Since this parᣊ�cular curve is a straight line, we know the slope (Δy/ Δx) of the line is constant. On this graph, the slope of the line represents acceleraᣊ�on. The steeper the curve (the line) is, the greater the acceleraᣊ�on. From the data presented here we know that the acceleraᣊ�on pictured in Figure 7 is constant.
Examining the curves on a velocity vs. ᣊ�me graph provides informaᣊ�on about the direcᣊ�on, speed and acceleraᣊ�on of the moving object. Take a look at the series of linear moᣊ�on graphs in Figure 7 and how they are interpreted.
Understanding the meaning of the shape and slope of a distance vs. ᣊ�me graph is essenᣊ�al to your understanding of linear moᣊ�on. Whether it's the speed or direcᣊ�on that is changing, the rate of acceleraᣊ�on will greatly influence the path of moᣊ�on. From driving your car to
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shooᣊ�ng a gun, these concepts apply to many situaᣊ�ons encountered in everyday life. Explore the world around you and see how many applicaᣊ�ons of linear moᣊ�on you can idenᣊ�fy!
Figure 7: Linear moᾷon graphs.
The Kinema뙕c Equa뙕ons
When you are at a red light and the light turns green, you hit the gas acceleraᣊ�ng your car up to the speed limit. During the period of ᣊ�me your foot is on the gas, your acceleraᣊ�on is close to constant. This is the case for many different types of moᣊ�on. From interpreᣊ�ng the 1‐D moᣊ�on graphs, acceleraᣊ�on is equal to the slope of velocity versus ᣊ�me graphs: a = dv/dt. Uᣊ�lizing calculus, the first kinemaᣊ�c equaᣊ�on can be derived using the following process:
(dv/dt) = a
(dv) = a ∙ dt
Integrate both sides of the equaᣊ�on.
∫ dv = ∫ a ∙ dt
∫ dv = a ∫ dt
If a is constant, it comes out of the integral.
v = at + C
The constant of integraᣊ�on is equal to the iniᣊ�al velocity at ᣊ�me zero. The resulᣊ�ng equaᣊ�on is the first kinemaᣊ�c equaᣊ�on.
vf = vo + at
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Another integraᣊ�on results in the next kinemaᣊ�c equaᣊ�on.
df = do + vot + ½at 2
These equaᣊ�ons are used to solve for the instantaneous posiᣊ�on, velocity, or acceleraᣊ�on of an object. They each relate two different variables and are summarized as the following:
df = d0 + v0t + ½ at 2 Relates distance and 뙕me
vf = v0 + at Relates velocity and 뙕me
vr 2= v0
2 + 2ad Relates velocity and displacement
where:
d is displacement which is equal to df ‐ d0 t is the amount of ᣊ�me the object is in moᣊ�on (s) vf is the final velocity (m/s) v0 is the iniᣊ�al velocity (m/s)
a is the acceleraᾷon (m/s2) d0 is the iniᾷal posiᾷon (m) df is the final posiᾷon (m))
Many equaᣊ�ons can be derived from the general equaᣊ�ons to find the above variables under different condiᣊ�ons as long as all variables except the one being solved for are known. Let's take a look at some examples. If the iniᣊ�al velocity of the object is equal to 0 m/s and iniᣊ�al posiᣊ�on is equal to 0 m, the kinemaᣊ�c equaᣊ�ons simplify to:
vf = at
df = ½ at 2
The following equaᣊ�on is useful to find the displacement an object travels when gravity is the only force acᣊ�ng on the object and the ᣊ�me is known. The acceleraᣊ�on of gravity, g, on earth is about 9.8 m/s2:
d = ½ gt2
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Figure 8: Skydivers experience free‐fall. To reduce the effects of air resistance, skydivers orient their bodies perpendicular to the ground. By doing this, they are able to reach a terminal velocity of about 120 mph!
Accelera뙕on and Gravity
Gravity causes objects to accelerate downward when falling. In reality, air resistance decreases the effects of gravity on the acceleraᣊ�ng objects unᣊ�l it balances the force of gravity and there is no longer a change in velocity — when the object reaches its terminal velocity (Figure 8). However, when we study linear moᣊ�on, it is convenient to neglect air resistance and focus only on the acceleraᣊ�on due to gravity (g), a state called freefall. In all cases of free fall, g (9.8 m/s2) is the acceleraᣊ�on due to the force of gravity that can act in a posiᣊ�ve way by increasing the speed of the object or negaᣊ�vely by decreasing the speed. Thus, for every second of free fall there is a change in velocity of approximately 10 m/s. This leads us to the equaᣊ�on for the acceleraᣊ�on of a falling object:
vf = gt
where v is the final velocity (m/s), g is the acceleraᣊ�on due to gravity (9.8 m/s2), and t is the elapsed ᣊ�me (s). Thus, velocity is a funcᣊ�on both of acceleraᣊ�on and of how long the object is acted on by that force.
Gravity is also responsible for slowing down an object thrown straight up into the air and acceleraᣊ�ng it back to Earth. In fact, the object reaches a point of zero velocity before it changes direcᣊ�ons. This characterisᣊ�c behavior is typical of all mass in the presence of gravity.
? Did You Know...
Physics is the study of how the world works. Currently, all physics has been described by the interacᾷon of parᾷcles in space at a certain ᾷme. We refer to this arena for parᾷcles to interact and collide with each other "space‐ ᾷme". Recently, physicist a new kind of geometric shape called the amplituhedron was discovered to simplify calculaᾷons for colliding parᾷcles. Could this be the end of "space‐ᾷme"?
Pictured to right: Amplituhedron
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2‐D Kinema뙕cs
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Projec뙕le mo뙕on takes into consideraᣊ�on objects that are moving in two direcᣊ�ons at the same ᣊ�me. Unlike linear moᣊ�on which only considers one direcᣊ�on, projecᣊ�le moᣊ�on acknowledges both horizontal and verᣊ�cal moᣊ�on. Think about two‐dimensional (2‐D) moᣊ�on as two, independent, one‐dimensional (1‐D) moᣊ�ons. The path of an object with projecᣊ�le moᣊ�on can be described as curved. This is where the concept of vectors comes into play.
Vectors
As you learned previously, a quanᣊ�ty that conveys informaᣊ�on about magnitude only is called a scalar quanᣊ�ty. Vectors, like velocity, describe magnitude and direcᣊ�on. Along with detailing informaᣊ�on about the path of moᣊ�on, vectors are also useful in physics because they can be separated into components.
An object with projecᣊ�le moᣊ�on has a velocity that can be represented by a diagonal vector. In Figure 9, the vector is labeled V. Vector V can be resolved (broken down) into an equivalent set of horizontal (x‐direcᣊ�on) and verᣊ�cal (y‐direcᣊ�on) components, which are at right angles to each other. The addiᣊ�on of the components results in the magnitude of the
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Figure 9: A vector can be broken down into horizontal and verᾷcal components.
vector.
Calcula뙕ng the x‐component of a Vector cosθ = Vx V
Vx = V (cosθ) Calcula뙕ng the y‐component of a Vector sinθ = Vy
V Vy = V (sinθ)
Calcula뙕ng Magnitude and Direc뙕on
Unit Vectors
Unit vectors are helpful when describing direcᣊ�on for problems that are in two or more dimensions. They provide informaᣊ�on about the components of a vector and have a magnitude of 1. The unit vector î is directed in the posiᣊ�ve x‐direcᣊ�on, and the unit vector ĵ is directed in the posiᣊ�ve y‐direcᣊ�on. To describe vectors by components, one can use unit vectors, as shown below.
Rx = Rxî
Rx describes the enᣊ�re x‐component of the vector, and Rx described the magnitude of the x‐ component.
Ry = Ryĵ
Combining the x‐ and y‐ components we can find the vector R:
R = Rx + Ry
R = Rxî + Ryĵ
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Figure 10: The path of a projecᾷle in the absence of air resistance is a perfect parabola. The horizontal component of velocity is 10 m/s throughout the
object's moᾷon.
Posi뙕on Vectors
To describe an object's locaᣊ�on in space, we use the posiᣊ�on vector. It is the vector that points from the origin to the object's posiᣊ�on in space at one instant. The posiᣊ�on vector can be wriĀen using unit vectors.
r = xî + yĵ
Just as we learned in 1‐D Kinemaᣊ�cs, an object's posiᣊ�on, average velocity, instantaneous velocity, average acceleraᣊ�on, and instantaneous acceleraᣊ�on are all important in describing an object's moᣊ�on through space. In 1‐D Kinemaᣊ�cs average velocity was found by taking the change in the object's displacement over a period of ᣊ�me. This is very similar to average velocity in 2‐D Kinemaᣊ�cs. Instead of using the change in the object's displacement, we use the change in the object's posiᣊ�on vector over a certain amount of ᣊ�me.
Instantaneous velocity in 1‐D Kinemaᣊ�cs was found by taking the derivaᣊ�ve of the object's displacement. For 2‐D Kinemaᣊ�cs, instead of displacement, we take the derivaᣊ�ve of the object's posiᣊ�on vector in 2 dimensions.
Average acceleraᣊ�on in 2‐D Kinemaᣊ�cs is virtually the same as it is in 1‐D Kinemaᣊ�cs. We take the change in the object's velocity vector over a certain amount of ᣊ�me.
Projec뙕le Mo뙕on
Typically, a projec뙕le is any object which, once projected, conᣊ�nues in moᣊ�on by its own inerᣊ�a and is influenced only by the downward force of gravity. This may seem counter‐intuiᣊ�ve since the object is moving both horizontally and verᣊ�cally, but gravity (an applied force) acts only on the verᣊ�cal moᣊ�on of the object. The term iner뙕a describes an object's resistance to external forces which could affect its moᣊ�on (both velocity or direcᣊ�onality). When there are no external forces acᣊ�ng upon an object, it will conᣊ�nue to travel in a straight line at a constant, linear velocity.
As shown in Figure 10, the projecᣊ�le with horizontal and verᣊ�cal moᣊ�on assumes a characterisᣊ�c parabolic trajectory due to the effects of gravity on the verᣊ�cal component of
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? Did You Know...
Cannons on Navy ba᧻le ships can fire at objects that are farther than fi뎌een miles away. The curvature of the Earth makes it impossible to see that far in the distance. That means that the cannons fire projecᾷles at objects they cannot physically see!
moᣊ�on. If air resistance is neglected, there are no horizontal forces acᣊ�ng upon the projecᣊ�le, and thus no horizontal acceleraᣊ�on. It might seem surprising, but a projecᣊ�le moves at the same horizontal speed no maĀer how long it falls!
Since projecᣊ�le moᣊ�on can be resolved into two independent direcᣊ�ons, 1‐D kinemaᣊ�c equaᣊ�ons can be applied to both components of the moᣊ�on separately. The kinemaᣊ�c equaᣊ�ons will allow you to solve for different aspects of a projecᣊ�le's flight, its height (verᣊ�cal), range (horizontal), and ᣊ�me of flight. Applying 1‐D kinemaᣊ�cs results in two sets of equaᣊ�ons for 2‐D moᣊ�on:
Height Range
yf = v0,yt ‐ 1/2gt 2 xf = v0,xt
vf,y = v0,y ‐ gt
vf,y 2= v0,y
2 ‐ 2gy
The x and y subscripts for velocity refer to the component of velocity in the x and y direcᣊ�on. These two sets of equaᣊ�ons (height and range) also incorporate ᣊ�me because the ᣊ�me of flight for the projecᣊ�le moᣊ�on is the same for both the verᣊ�cal and horizontal moᣊ�ons. Noᣊ�ce that there is only one equaᣊ�on for range, while there are three equaᣊ�ons for height! This is due to the fact there is no acceleraᣊ�on in the horizontal direcᣊ�on. You may oĀen be required to find the ᣊ�me of flight using the height equaᣊ�ons in order to determine the range of the projecᣊ�le.
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Figure 11: With air resistance, the trajectory looks like a "squashed" parabola, and the range of the object's
flight is noᾷceably affected.
Launch Angle
If the projecᣊ�le is fired at an angle, the range is a funcᣊ�on of the iniᣊ�al launch angle, θ, the launch velocity and the force of gravity. Using algebra, you can derive the following expression from the kinemaᣊ�cs equaᣊ�ons:
R = v2 sin(2θ) g
It is important to remember that in many cases, air resistance is not negligible (Figure 11) and affects both the horizontal and verᣊ�cal components of velocity. When the effect of air resistance is significant, the range of the projecᣊ�le is reduced and the path the projecᣊ�le follows is not a true parabola.
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Circular Mo뙕on
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Figure 12: Diagram of the relaᾷonship between ᾷme period, velocity and radius during circular moᾷon.
Imagine you are driving in a car with the steering wheel turned so that the car follows the path of a perfect circle. If the speedometer read the same speed the whole ᣊ�me, the moᣊ�on of the car would be described as uniform circular mo뙕on.
Uniform Circular Mo뙕on
Uniform circular moᣊ�on is the moᣊ�on of an object in a circle with constant speed, when the object covers the same distance in each instant of ᣊ�me.
Recall that the average speed is equal to the distance traveled divided by the ᣊ�me. For one revoluᣊ�on in a circle the distance traveled is equal to the circumference of a circle. The ᣊ�me to make one revoluᣊ�on is called a period. The equaᣊ�on to find the average speed of an object traveling in uniform circular moᣊ�on is:
Vavg = 2πr T
where T is the ᣊ�me period, and r is the radius of the circular path the object travels. As the radius increases, velocity increases. For example, imagine four spherical objects moving in a uniform circular path around a staᣊ�onary point (Figure 12). Each ball has the same ᣊ�me period because they move around the circle in the same amount of ᣊ�me, but the speed at
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Figure 13: The change in velocity (Δv) is shown in the small diagram to the right, and is the sum of the two velocity vectors v1 and v2.
which each ball moves differs. As the radius gets larger, the speed also increase in order to cover greater distance in the same amount of ᣊ�me as the other objects.
? Did You Know...
The word centripetal stems from the La뙕n meaning "toward center".
Centripetal Accelera뙕on
Newton's First Law states that an object in moᣊ�on will stay in moᣊ�on unless acted on by an external force. Figure 13 depicts an object moving in a circular moᣊ�on at a constant speed. We see that the direcᣊ�on of the velocity vector is changing as the object travels around the circular path. Therefore, there must be an acceleraᣊ�on causing the mass to change direcᣊ�on. What is the direcᣊ�on of this acceleraᣊ�on? Average acceleraᣊ�on is calculated by dividing the
change in velocity by ᣊ�me, so the acceleraᣊ�on is in the same direcᣊ�on as the change in velocity which is toward the center of the circle! This acceleraᣊ�on is called centripetal accelera뙕on. No maĀer what two velocity vectors you choose, the acceleraᣊ�on vector is always perpendicular to the tangenᣊ�al velocity toward the center of rotaᣊ�on. Thus, an object in circular moᣊ�on can always be thought of as acceleraᣊ�ng toward the center of the circle, even though the radius of rotaᣊ�on remains constant. The magnitude of centripetal acceleraᣊ�on can be expressed in terms of the linear velocity and the radius of rotaᣊ�on:
ac = v 2
R
The force causing centripetal acceleraᣊ�on is just large enough to keep the object in its circular path. This centripetal force for uniform circular moᣊ�on alters the direcᣊ�on the object is traveling in, but not the speed. Just like linear moᣊ�on, force is a measure of the mass in moᣊ�on mulᣊ�plied by its acceleraᣊ�on. The magnitude of the force is related to the acceleraᣊ�on of the object through the following relaᣊ�onships based on Newton's Second Law of Moᣊ�on:
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Figure 14: Swings at an amusement park exhibit a circular path of moᾷon (white line).
Figure 15: Banked turns keep cyclists from being pushed off a
track.
Let's analyze a real world example of centripetal moᣊ�on. The seats on a swinging ride at an amusement park twirl around the rotaᣊ�ng central pole, and their moᣊ�on is described as a circle (Figure 14). The seats are connected to a chain, which provides an (inward) centripetal force to keep the swings from flying off in a straight line. Increasing the velocity of rotaᣊ�on forces the chain out at a wider angle to the point where the horizontal component of its tension provides the necessary centripetal force inward. Meanwhile, the verᣊ�cal component of the tension must balance out the force of gravity downward.
If you have ever been on a rotaᣊ�ng merry‐go‐round, you have probably felt as if something is pulling toward the outside edges, forcing you to hold onto the bar to keep you from falling off. A common word that is oĀen used when discussing circular moᣊ�on is centrifugal force (center fleeing), which is the "ficᣊ�ᣊ�ous" force that seems to push a rotaᣊ�ng object away from the center of rotaᣊ�on. The aspects of this "force" are really a misconcepᣊ�on: what you are feeling is the centripetal force exerted by the bar to keep you from traveling in a direcᣊ�on tangent to the rotaᣊ�on. The centrifugal force is "felt" when the bar pulls you inward, giving you the impression that a force is pushing you outward. If you let go of the merry‐go‐round, your inerᣊ�a (remember Newton's First Law) will keep you moving in a direcᣊ�on tangent to the rotaᣊ�on (which is a straight line in the direcᣊ�on of your velocity at that very moment), not outward. To prevent this from occurring during sports
involving circular tracks and even on highways, the turns are banked or inclined. This bank provides a normal force on the object, which is perpendicular to the track surface. The horizontal component of the normal force plus the horizontal component of the force of fricᣊ�on between the ᣊ�res and track are the source of centripetal force inward. If the angle of the track and the speed of the vehicle are just right, no sideways fricᣊ�on forces are required to turn on the track (Figure 15).
Angular Mo뙕on
Consider a mass moving at a constant speed in a circular path. Since the mass is not moving in a straight line, its posiᣊ�on is described as the angle, θ, from the x‐axis or angular posi뙕on (Figure 16). Angular posiᣊ�on is
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Figure 16: A dark gray mass is moving in a circular path at radius, R, and has
traveled an arc length of l.
expressed in radians (instead of degrees) and is related to arc length, l, and the radius, R, of a circular path.
Rθ(rad) = l
Angular posiᣊ�on is posiᣊ�ve when measured counterclockwise and negaᣊ�ve when measured clockwise.
The arc length of a complete circle is equal to the circumference (2Πr) and there are 360° in a complete circle. Therefore, you can convert back and forth between degrees and radians using the equaᣊ�on:
360° = 2πr rad
In linear moᣊ�on, velocity is an object's change in posiᣊ�on divided by the ᣊ�me the object is in moᣊ�on:
△ d / △t
In circular moᣊ�on, angular velocity is the amount of rotaᣊ�on or revoluᣊ�on per unit of ᣊ�me. Average angular velocity, ωavg, is the number of radians per second (rad/s) over a period of ᣊ�me and instantaneous angular velocity, ω, is the number or radians over an infinitesimal period of ᣊ�me:
Average: ωavg = △ϑ / △t
Instantaneous: ω = dθ / dt
Average angular velocity can also be calculated by measuring the period of mo뙕on, P, which is the amount of ᣊ�me it takes an object to make one full revoluᣊ�on. Angular velocity is mathemaᣊ�cally related to period as:
ωavg = 2πr / P
When an object rotates at a fixed radius, R, its linear velocity at a given instant is always tangent to its circle of rotaᣊ�on. For this reason, the term tangenᣊ�al velocity refers to the linear velocity of a rotaᣊ�ng object (as opposed to angular velocity). A tangent line is one that touches the circle at one point, but never intersects it. Therefore, the direcᣊ�on of the tangenᣊ�al velocity is always changing while the object makes a revoluᣊ�on. Since the linear distance traveled around one revoluᣊ�on is the circumference, the tangenᣊ�al velocity of an object in uniform circular moᣊ�on is the total distance, 2Πr, over the ᣊ�me period, P:
v = 2πr / P = Rω
An object freed from its rotaᣊ�on will conᣊ�nue traveling in a straight line tangent to the previous circle of rotaᣊ�on because the linear velocity is always tangent the circle.
In the case where the angular velocity is not constant, or non‐uniform circular mo뙕on, the angular acceleraᣊ�on, (rad/s2), is the rate of change of the angular velocity, ω:
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α = dω / dt
The linear kinemaᣊ�cs equaᣊ�ons easily translate into rotaᣊ�onal kinemaᣊ�cs equaᣊ�on when linear variables are replaced with rotaᣊ�onal ones:
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Lab 2/Lab2Kinematics.docx - Copy.docx
Lab 2 Kinematics PHY250L”
Student Name:
Kit Code (located on the lid of your lab kit): AC-65CS487
“Pre-Lab Questions”
1. “What is the acceleration of a ball that is vertically tossed up when it reached its maximum height?”
0
“
2. “The displacement of a particle as it varies with time is given by the equation
x(t) = (10.0 m/s) t + (2.50 m/s²) t². Find the particle’s instantaneous velocity and instantaneous acceleration at t = 4.00 seconds.”
Click here to enter text.
3. “What does a positive and negative slope represent for a velocity vs. time graph?” This exemplifies an increase when positive and a decrease when negative of an object's velocity in respect to time.
5. “Derive the second kinematic equation by integration of the first kinematic equation. Then derive the third kinematic equation by using algebra to combine the first and second kinematic equations.” Click here to enter text.
“Experiment 1: Distance of Free Fall”
“Table 1: Washer Free Fall Data”
|
“Trail” |
“Drop Height (m)” |
“Time (s)” |
|
“1” |
4.2 Meters |
1.0 |
|
“2” |
|
.88 |
|
“3” |
|
.95 |
|
“Average” |
|
.9433 |
“Description of Auditory Observations of Equally Spaced Hex Nuts:” Consistent sound
“Description of Auditory Observations of Unequally Spaced Hex Nuts:” Inconsistent sound
“Post-Lab Questions”
1. “Record your hypothesis from Step 1 here. Use evidence from your results to explain if your hypothesis was supported or not.”
Varying distances will have a direct correlation with time. The shorter the distance the smaller the value of time will be. Vise versa also applies, the longer the distance, the larger the value of time will be. I have observed this during the first exercise.
2. “What was the difference between the noise patterns for equally spaced hex nuts compared to the unequally spaced hex nuts?”
The equally spaced hex nuts gave off a consistent sound pattern, while the inconsistently spaced hex nuts were the opposite, and let off a sporadic sound pattern.
3. “If the noise patterns were different, explain why. If they were similar, explain why.” The sound was different on the unequally spaced nuts, due to the less distance in between each nut. This cause a shorter time period in which there wasn’t any sound produced, until the next nut hit the floor.
4. “Using the time it took a single hex nut to reach the pan, calculate the height from which it was dropped. Is this accurate compared to your known height? Explain your conclusion.”
The time in which I had used the stopwatch to calculate the drop distance was 1 second flat. This gave me a calculation of 4.9 Meters, or roughly 16 ft. In comparison to the actual measurement of 13.9 ft or 4.2 meters, this is very close.
5. “A student ran this experiment, and instead of dropping the hex nut, he threw it. This gave the nut a velocity of v(t) = (12 m/s²)t + 5 m/s. What is the hex nut’s displacement as a function of time if its position at t = 0 seconds is 0 meters? If the hex nut dropped 1.2 meters, how long did it take for the nut to reach the ground?” Click here to enter text.
“Experiment 2: Distance Traveled by a Projectile”
“Pre-Lab Questions”
1. “In one of your experiments, you will roll a marble down a ramp to provide an initial horizontal velocity. Suppose you start the marble at rest (v˳= 0 m/s) and it travels a distance of, d, down the ramp. Use 1-D kinematics to predict the velocity of the ball (vᶠ) at the bottom of the ramp. Hint: the acceleration of the ball down the ramp is 9.81*sin(θ) m/s² where θ is the angle of the ramp. Record your answer in variables (you will calculate the velocity with magnitudes when you perform the experiment).”
Click here to enter text.
2. “Use the kinematic equations to derive a general equation for the time it takes a ball dropped from rest at vertical height, h, to reach the ground. Use this to write a general equation for the distance travelled by a projectile that is rolling off a table of height, h, with a horizontal speed of V˳ₓ.” Click here to enter text.
3. “Prove that launching a projectile at 45° provides the largest range. Write the range as a function of θ. Take the derivative of the range with respect to θ and find the maximum angle.”
Click here to enter text.
4. “A butterfly flies along with a velocity vector given by v = (a-bt²) Î + (ct) ĵ where a=1.4 m/s, b=6.2 m/s³, and c=2.2 m/s². When t= 0 seconds, the butterfly is located at the origin. Calculate the butterfly’s position vector and acceleration vector as functions of time. What is the y-coordinate as it flies over x = 0 meters after t = 0 seconds?”
Click here to enter text.
“Data:”
“Table 1: Range and Velocity of Projectile at Ramp Distance 1”
|
“Ramp Incline (degrees):” 25 |
|
|
“Ramp Distance (m):”33.5 |
|
|
“Trial” |
“Measured Distance (m)” |
|
“1” |
42.5 |
|
“2” |
41.5 |
|
“3” |
41. |
|
“4” |
42.25 |
|
“Average” |
41.8 |
“Table 2: Range and Velocity of Projectile at Ramp Distance 2”
|
“Ramp Distance (m):” 28.5 |
|
|
“Trial” |
“Measured Distance (m)” |
|
“1” |
35.8 |
|
“2” |
37 |
|
“3” |
37.6 |
|
“4” |
36.5 |
|
“Average” |
36.7 |
“Table 3: Range and Velocity of Projectile at Ramp Distance 3”
|
“Ramp Distance (m): “23.5 |
|
|
“Trial” |
“Measured Distance (m)” |
|
“1” |
32.5 |
|
“2” |
32.5 |
|
“3” |
33.3 |
|
“4” |
33.5 |
|
“Average” |
32.95 |
“Post-Lab Questions:”
5. “Use your predictions of velocity and range from the Pre-Lab Questions and the data recorded from your experiment to complete Table 4.”
“Table 4: Velocity and Range Data for all Ramp Distances”
|
Ramp Distance (m) |
Calculated velocity (m/s) |
Predicted Range (m) |
Average Actual Range (m) |
Percent Error |
|
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
|
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
|
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
6. “How do your predictions compare to the observed data? Explain at least two reasons for differences.”
Click here to enter text.
7. “If you were to fire a paintball pellet horizontally and at the same time drop the same type of paintball pellet you fired from the paintball gun, which pellet would hit the ground first and why is this so?” They would both come into contact with the ground at the same exact time. They both experience the same amount of downward air resistance, therefore they both are approaching the ground at the same exact rate.
8. “A marble slides down a wacky ramp with a velocity given by v = (0.5t²-3.0t) Î + (0.33t³-0.6t) ĵ. At t = 3 seconds the particle is shot off the ramp and behaves like a projectile. What is the magnitude of the velocity of the marble when it leaves the ramp? What is the marble’s acceleration vector while it is on the ramp? If the marble falls for 6 seconds when it is shot off the ramp at a 45° angle, what will be its displacement in the x-direction?” Click here to enter text.
“Insert photo of your experimental setup with your name clearly visible in the background:”
“Experiment 3: Squeeze Rocket™ Projectiles”
“Table 5: Projectile Data for Rockets and Different Launch Angles”
|
Launch Velocity (m/s) |
Initial Angle |
Time (s) |
Average Time (s) |
Predicted Range (m) |
Actual Range (m) |
Average Range (m) |
Range % Error |
|
0.5
|
“90°” |
1.34
|
1.42
|
1
|
.67
|
.905
|
17
|
|
|
“90°” |
1.41 |
|
1
|
.83
|
|
33
|
|
|
“90°” |
1.53 |
|
1
|
.99
|
|
1
|
|
|
“90°” |
1.38 |
|
1
|
1.1
|
|
10
|
|
|
80 |
1.4 |
1.56
|
1.5
|
1.2
|
1.37
|
30
|
|
|
80 |
1.6 |
|
1.5
|
1.4
|
|
10
|
|
|
80 |
1.51 |
|
1.5
|
1.37
|
|
13
|
|
|
80 |
1.72 |
|
1.5
|
1.52
|
|
2
|
|
|
25 |
1.05 |
1.04
|
3 |
3.09
|
3.02
|
9
|
|
|
25 |
1.01 |
|
3
|
3.25
|
|
25
|
|
|
25 |
1.11 |
|
3
|
2.81
|
|
19
|
|
|
25 |
.98 |
|
3
|
2.92
|
|
8
|
|
|
1.28 |
1.05 |
1.25
|
2
|
1.98
|
2.2
|
2
|
|
|
60 |
1.15 |
|
2
|
2.48
|
|
48
|
|
|
60 |
1.22 |
|
2
|
2.25
|
|
25
|
|
|
60 |
1.36 |
|
2
|
2.1 |
|
10
|
“Post-Lab Questions:”
1. “Which angle provides the greatest range? Which provides the least? Based on your results, which angle should give you the greatest ranged for projectile motion?”
The lower the angle the further the range as exemplified by the data. The angle at which the projectile is launched has a direct result on the distance at which it will travel.
2. “What role does air resistance play in affecting your data?”
Air resistance plays a huge role in the above data because of the mass of the object itself. If the projectile had been more dense I wouldn’t see the projectile sidewind as much as it did, however it would take a bit more initial force to launch the projectile.
3. “Discuss any additional sources of error, and suggest how these errors could be reduced if you were to redesign the experiment.”
As stated above I would use an object more dense in structure, due to the motion at which the projectile traveled during the experiment this leaves room for some inconsistencies.
4. “How could kickers on a football team use their knowledge of physics to better their game? List at least two other examples in sports or other applications where this information would be important or useful.”
A kicker on a football game could use knowledge of what angle to kick the football at to make a precise kick in the field goal, this way he could find exactly at what power and angle to kick the ball at for the desired distance to the field goal. Another practical use would be in golf. Having the ability to gauge power, and angle for the desired distance in order make it onto the putting green, or even a hole in one could be powerful.
5. “A student buys a high powered toy rocket gun and tries to do this experiment. He decides to shoot the first rocket with an initial angle of 0 degrees. The student knows that the rocket has an initial velocity of 5 m/s when he shoots it off of the 3 meter high table. The toy gun is able to give the rocket a horizontal acceleration of (2.1 m/s³)t in the same direction as the initial velocity. The vertical acceleration, directed downwards, is g. Assume air resistance can be ignored. What is the horizontal displacement of the student’s toy rocket?”
3 meters
“Experiment 4: Balancing Centripetal Force”
“Pre-Lab Questions:”
1. “In this lab, you will be rotating a mass on one side of a string that is balanced by a second mass on the other end of the string (Figure 5). Apply Newton’s Second Law of Motion to mass 1, m₁, and mass 2, m₂, to solve for the period of mass 1.
Hint: assume m₁ = 4m₂. How is the centripetal force on m₁ related to the force of gravity on m₂.”
The centripetal force negates and counteracts the forces of gravity on the end with more washers.
2. “Draw a free body diagram and solve for the centripetal acceleration in terms of θ and g for one person riding on the amusement ride in Figure 3.”
Click here to enter text.
3. “The wheel of fortune is 2.6 meters in diameter. A contestant gives the wheel an initial velocity of 2 m/s. After rotating 540 degrees, the wheel comes to a stop. What is the angular acceleration of the wheel?”
Click here to enter text.
4. “The angle that a spoke on a bicycle wheel has rotated behaves according to function θ(t) = at² + bt where a = 0.6 rad/s² and b = 0.3 rad/s. Find the angular velocity of the spoke as a function of time and the angular acceleration as a function of time. Then find the instantaneous angular velocity and the instantaneous angular acceleration at t = 3 seconds.”
Click here to enter text.
“Data:”
“Table 1: Rotational Data”
|
Radius (m) |
Time per 15 revolutions (s) |
“Period (s)” |
Expected Value |
Percent Error (%) |
|
“0.25” |
5.96 |
.397 |
.5 |
10.3 |
|
“0.40” |
7.6 |
.51 |
.75 |
24 |
|
“0.15” |
4.94 |
.33 |
.4 |
7 |
“Post-Lab Questions:”
5. “Compare your measured data to your predicted values with a percent error calculation. Explain any difference with an error analysis.”
Click here to enter text.
6. “Draw a circle to represent the path taken by your rotating mass. Place a dot on the circle to represent your rotating washer. Add a straight line from the dot to the center of the circle, representing the radius of rotation (the string). Now label the direction of the tangential velocity and the centripetal force.” Insert photo of the circle with your name clearly visible in the background:
7. “Use your data to calculate the average velocity, angular velocity, and centripetal acceleration for the mass in each radius.”
Click here to enter text.
8. “Refer to the picture in Figure 3 again (picture in Pre-Lab question #2). Before the apparatus begins to spin, the wires connecting the swings to the top of the structure will be completely vertical. Once the apparatus begins to spin the swings move outward radially, but also upwards vertically. From where does the force causing this vertical acceleration come?”
Click here to enter text.
9. “Refer to the picture in Figure 3. Imagine that the swings are rotating around the center with a constant speed, and the wire connecting the swings to the center pole is at a 45 degree angle. The angular velocity of the center pole is then doubled. Does this mean that the chairs’ velocities will increase by a factor of two, less than two, or more than two? Explain your reasoning.”
Click here to enter text.
10. “A uniform disc is rotating around a frictionless, vertical axle that passes through its center has as radius of R=0.300m and a mass of 25.0 kg. The disc rotates according to θ(t) = (2.20 rad/s²) t² + (5.63 rad/s) t. When the wheel has rotated 0.200 rev, what is the resultant linear acceleration of any point on the disc?”
Click here to enter text.
Lab 3/Lab 3- Newton's Laws.pdf
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Learning Objecves
Formulate the law of inera Relate force and acceleraon Apply acon and reacon pairs to forces Draw and explain free body diagrams Apply Newton's 2nd Law to the Atwood Machine
Introducon
The laws of physics that we know today were discovered and have been studied for centuries. In the fourth century B.C. Aristotle proposed the general belief that a force causes a constant velocity. In addion to Aristotle, Isaac Newton, famous for his Laws of Moon, based his work on the discoveries and experiments of Galileo and Johannes Kepler. In order to reduce the plague from spreading through the college where Newton was a fellow, it was temporarily closed. During this closure, Newton spent a few years in relave isolaon where he started to
1/4/2017 Lab 3: Newton's Laws
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Figure 1: The moĕon of moon as predicted by Newton's First Law should follow the doĥed path. Since the moon does not
follow the predicted path, there must be a force acĕng on it.
formulate ideas on mathemacs and physics. Newton was trying to figure out what objects in orbit and objects falling toward the Earth have in common. The answer actually required a new type of math: calculus! Newton formulated quantave explanaons for the moon of falling objects, orbing objects, pulley systems, and much more. These ideas were able to explain all types of moon and can be broken down into three basic laws.
Force
While force can be described as a push or a pull, it is more clearly defined as an acon that causes an object to change its moon. Force is what causes the direcon of an object's velocity to change. Therefore, a net force acng on an object causes acceleraon.
Newton's First Law of Moon
Newton's First Law of Moon states that an object will maintain its state of moon unl acted upon by an external force. In other words, an object will remain at rest or move at a constant velocity unl an outside force acts upon it (Figure 1). Newton's First Law is also called the Law of Inera. Inera is an object's tendency to resist changes in moon (speed or direcon). Ma台�er has this property whether it is at rest or in moon. When a net force on an object is applied, the object will accelerate in the direcon of that force. If the net force on the object equals zero then there will not be a change in its acceleraon. When discussing net force, the vector sum of the forces, the resultant, must be examined.
The movement of planets around the sun is an example of inera. Planets have a lot of mass, and therefore a great amount of inera ‐ it takes a huge force to accelerate a planet in a new direcon. The pull of gravity from the sun keeps the planets in orbit; if the sun were to suddenly disappear, the planets would connue at a constant speed in a straight line, shoong off into space!
Vector Addion
Unlike addion of scalars, vector addion requires a geometrical process where the tail of the second vector is placed at the p of the first vector, o�en referred to as the tail to p method. If an object is acted upon by force A and then by force B, it is same as the object undergoing a single force C (Figure 2). This value is known as the resultant or vector sum.
A + B = C
1/4/2017 Lab 3: Newton's Laws
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Newton also noted that the greater an object's mass, the more it resisted changes in moon. Therefore, he concluded that mass and inera are directly proporonal: when mass increases, inera increases. This predicon produced Newton's Second Law of Moon, an expression for how an object will accelerate based on its mass and the net force applied to the object. This law can be summarized by the equaon:
ΣF = ma
where ΣF is the sum of all forces acng on the object, m is its mass and a is its acceleraon. The kilogram (kg) is the standard measurement for mass, and meter/second/second (m/s2) is the standard measurement for acceleraon. The standard measurement for force is the Newton, where 1 N = 1 kg∙m/s2. Comparing this equaon to the first one (w = mg) helps reinforce the difference between mass and force (such as weight).
Free body diagrams (as seen in Figure 3) are a useful tool for solving problems related to Newton's Second Law of Moon. They allow you to idenfy and draw all of the forces acng on an object. Objects can be represented by simple shapes like circles and squares. Forces are represented by solid arrows. It is helpful to visualize the net force acng on an object by labeling all objects and forces included in the free body diagram.
Since force and acceleraon are both vector quanes, Newton's Second Law can be wri台�en as a vector equaon.
ΣF = ma
It can also be broken up into x‐ and y‐ components (for three dimensional problems there is also a z‐component).
ΣFx = max
ΣFy = may
Figure 3: Free body diagram of a block hanging form a beam by a string.
Newton's Third Law of Moon
Newton's Third Law of Moon states that for every acon (force) there is an equal and
1/4/2017 Lab 3: Newton's Laws
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Figure 3: Newton's Third Law of Moĕon explains why you move backwards when you throw a ball on ice.
opposed reacon. Imagine standing on an ice skang rink and holding a ball (Figure 3). Inially you are not moving. You throw a ball and as you do, you start to move backwards on the ice. When you exerted a force on the ball it exerted a force on you equal in magnitude and opposite in direcon. Even when you walk, you push against the ground, and it pushes right back!
Newton's three laws of moon govern the relaonship of forces and acceleraon. There are many applicaons of Newton's Laws in your everyday life. To get that last bit of ketchup from the bo台�le, you shake the bo台�le upside‐ down, and quickly stop it (with the lid). Consider riding in a car. Have you ever experienced inera while rapidly accelerang? Thousands of lives are saved every year by seatbelts, which are safety restraints that protect against the inera that propels a person forward when a car comes to a quick stop.
© 2014 eScience Labs, LLC.All Rights Reserved
Lab 3/Lab3NewtonsLaws.docx - Copy.docx
Lab 3 Newton’s Laws PHY250L
Student Name: Click here to enter text.
Kit Code (located on the lid of your lab kit): Click here to enter text.
“Pre-Lab Questions”
“Use the free body diagram of the pulley (Figure 5) to answer the Pre-Lab Questions.”
1. “Draw free body diagrams for M₁ and M₂.”
“Insert photo of diagram with your name clearly visible in the background:”
2. “Apply Newton’s Second Law to write the equations for M₁ and M₂. You should get two equations with tension in the string, weight for each mass and accelerations for each mass (a₁ and a₂).”
3. “Post-Lab Question 2 results in two equations with three unknowns! A third equation is required to solve the system. What is the third equation?”
Click here to enter text.
4. “In the diagram to the right, there is a 5 kg block of ice on a slippery slope positioned 25° from horizontal. What is the acceleration of the block as it slides down the ramp?”
Click here to enter text.
5. “An object is acted upon by a force that can be modeled by F(t) = 5.6 N î + 2.4 N ĵ. The object has a mass of 3.0 kg and starts at rest. Calculate the velocity v(t) as a function of time.”
Click here to enter text.
“Experiment 1: Graphing Linear Motion”
“Table 1: Motion of Water Observations”
|
“Motion” |
“Observations” |
|
“a” |
water sloshed to one side |
|
“b” |
Water eventually remained at rest |
|
“c” |
Water sloshed to the left and then to the right |
|
“d” |
water sloshed forwards |
“Table 2: Observations After Flicking Notecard Off of Cup”
|
“Trial” |
“Observations” |
|
“1” |
Washer dropped to the bottom of the cup |
|
“2” |
Washer dropped to the bottom of the cup |
|
“3” |
Washer dropped to the bottom of the cup |
|
“4” |
Washer dropped to the bottom of the cup |
|
“5” |
Washer dropped to the bottom of the cup |
“Post-Lab Questions”
1. “Explain how your observations of the water and washer demonstrate Newton’s law of inertia.”
For the washer, the only force acting upon it is gravity, and friction. The friction moved the washer slightly in the direction the index card flew, but gravity overtook and brought the washer to the bottom of the cup. For the water, the only forces acting on the body of water was gravity and the edges of the container.
2. “Draw a free body diagram of your containers of water from the situation in Part 1 Step 4d. Draw arrows for the force of gravity, the normal force (your hand pushing up on the container), and the stopping force (your hand accelerating the container as you stop). What is the direction of the water’s acceleration?” Forward
“Insert photo of diagram with your name clearly visible in the background:”
3. “Can you think of any instances when you are driving or riding in a car that is similar to this experiment? Describe two instances where you feel forces in a car in terms of inertia.”
When the brakes are applied abruptly. Your body remains in motion as the vehicle slows down causing you to feel constrained when the seatbelt slows your body down.
4. “You and your friend both preform 3b of Part 1 of this experiment. You walk at 0.5 m/s, but your friend walks at 1.5 m/s. Which container of water will experience a greater net force?”
My friends container will experience greater net force due to the higher amount of intertia
“Experiment 2: Newton’s Third Law and Force Pairs”
“Table 3: Force on Stationary Springs”
|
“Force on Stationary 10 N Spring Scale (N)” |
|
5 |
|
“Force on Stationary 5N Spring Scale (N)” |
|
5 |
“Table 4: Spring Scale Force Data”
|
“Suspension Set Up” |
“Force (N) on 10 N Spring Scale” |
“Force (N) on 5 N Spring Scale” |
|
“0.5 kg Mass on 10 N Spring Scale” |
.5 |
|
|
“0.5 kg Mass with String on 10 N Spring Scale” |
.5 |
|
|
“0.5 kg mass, string and 5 N Spring Scale on 10 N spring scale” |
.8 |
.7 |
|
“0.5 kg mass, string and 5 N Spring Scale on 10 N spring scale on Pulley” |
.6 |
.7 |
“Post-Lab Questions”
1. “How did the magnitude of the forces on both spring scales compare after you moved the 10 N spring scale?”
The forces acted upon each scale were the same.
2. “How did the magnitude of the forces on both spring scales compare after you move the 5 N spring scale?”
The forces acted upon each scale were the same.
3. “Use Newton’s Third Law to explain your observations in Questions 1 and 2.”
For each action there is an equal and opposite reaction. So each force applied reacted to each scale equally.
4. “Compare the force on the 10 N spring scale when it was directly attached to the 0.5 kg mass and when there was a string between them.”
there was no difference in the amount of force, however there was a slight increase due to the mass of the string itself.
5. “Compare the force on two spring scales in Steps 5 and 6. What can you conclude about the tension in a strong?”
Tension on a string will eventually reach the point where it’s taught, and hold the amount of weight that is applied to it.
6. “Olympic spring Usain Bolt set the world record in the 100 meter dash with a time of 9.58 seconds. The physics behind his performance are impressive. Physicists have modeled his position as a function of time to be as follows:”
“where A=110 m/s, B=12.2 m/s, k=0.9 1/s. Using this information, find his acceleration when t = 0 seconds. If his mass was 86 kg at the time of the race, what was his force at t = 0 seconds. What was the force from the ground on Bolt at t = 0 seconds?”
Click here to enter text.
“Insert photo of your experimental setup with your name clearly visible in the background:”
“Experiment 3: Newton’s Third Law and Force Pairs”
“Table 5: Motion Data”
|
“Mass of 15 Washers (kg)” |
.040 |
“Average of Mass of Washer (kg)” |
.00266 |
|
“Procedure 1” |
|||
|
“Height (m):” .365 |
|||
|
“Trial” |
“Time(s)” |
||
|
“1” |
1.25 |
||
|
“2” |
1.13 |
||
|
“3” |
1.09 |
||
|
“4” |
1.06 |
||
|
“5” |
1.31 |
||
|
“Average” |
1.17 |
||
|
“Average Acceleration (m/s²)” |
.266 |
||
|
“Procedure 2” |
|||
|
“Height (m):” .365 |
|||
|
“Trial” |
“Time(s)” |
||
|
“1” |
.78 |
||
|
“2” |
.69 |
||
|
“3” |
.66 |
||
|
“4” |
.72 |
||
|
“5” |
.65 |
||
|
“Average” |
.7 |
||
|
“Average Acceleration (m/s²)” |
.744 |
“Post-Lab Questions”
1. “Draw a free body diagram for M₁ and M₂ in Procedure 2. Draw force arrows for the force due to gravity acting on both masses (Fᵍ₁ and Fᵍ₂) and the force of tension (Fᵀ). Also draw arrows indicating the direction of acceleration, a.”
“Insert photo of diagram with your name clearly visible in the background:”
2. “Use Newton’s Second Law to write an equation for each of the free body diagrams you drew in Question 1. Be sure to use the correct signs to agree with your drawings. Solve these four equations for the force of tension (Fᵀ). Your answer should be written in variable form.”
Click here to enter text.
3. “Set the two resulting expressions for the force of tension equal to one another (as long as the string does not stretch, the magnitude of the acceleration in each equation is the same). Replace Fᵍ₁ and Fᵍ₂ with M₁ and M₂, respectively. Solve the resulting equation for a. Then, go back to Questions 2 and solve for the Fᵀ.”
Click here to enter text.
4. “Calculate the acceleration for the two sets of data you recorded and compare these values to those obtained by measuring distance and time using percent error. What factors may cause discrepancies between the two values?”
Click here to enter text.
5. “Calculate the tension in the string for the falling washers. From these two values, and the one where the masses were equal, what trend do you observe in the tension in the string as the acceleration increases? Show all calculations.”
Click here to enter text.
6. “Using the theoretical acceleration found in Question 4 for Procedure 1, find the velocity of the block as a function of time by integration.”
Click here to enter text.
Lab 4/Lab 4- Friction.pdf
1/4/2017 Lab 4: Friction
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Learning Objecves
Explore and explain the difference between stac and kinec fricon Determine the dependence of the force of fricon on the normal force Apply the force of fricon to objects on an incline
Introducon
Newton's Laws of Moon can help idenfy forces acng on objects that can not be seen with the unaided eye and would go unnoced without them. Friconal force and normal force are two forces that can be inferred from an object's moon or lack thereof.
Normal Force
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Figure 1: Free body diagram of an object at rest.
Figure 2: Surfaces have microscopic imperfec笕�ons that contribute to fric笕�on.
When you place a book on a table, why doesn't it fall through the table? The force due to gravity (Fg = mg) never stops pulling down on the object, yet it remains staonary on the table. Newton's Second Law states the net force on an object must be zero if the object is at rest. Therefore, there must be another force acng on the object that is opposing the force of gravity. We call this force the normal force (Figure 1). Similar to tension, normal force is also passive and will only push back as hard as the applied force. Another important characterisc is that the normal force only pushes (applies a force) perpendicular to its surface.
Fricon
Newton's First Law states that an object in moon will remain in that state of moon unl an outside force acts on it. If you roll a ball across the floor or slide a book across a table, you will see them eventually come to rest. Similar to normal force and gravity, there seems to be another force at play that stops the rolling ball. The force that opposes the
direcon of moon and causes objects to come to rest is called friconal force. Fricon exits between two solid surfaces no maᘀer how smooth they may seem to the naked eye. If you zoom in on the surfaces, you will observe microscopic bumps that impede the moon of an object (Figure 2). Unlike the normal force, the friconal force always acts parallel to the surface.
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Stac vs. Kinec Fricon
An object does not always move when you exert a horizontal force. For example, have you ever tried pushing a car that has broken down and not been able to get it in moon? Stac fricon keeps an object at rest. If you push with a greater force and the mass sll has not changed posion, stac fricon increases. Once you push hard enough to put the car in moon, you have overcome the maximum amount that stac fricon can push back. Stac fricon can be modeled by the inequality Ff ≤ μsFN, where Ff is the force of fricon, μs is the coefficient of stac fricon, and FN is the normal force. The less than or equal to sign indicates that the force of stac fricon can range from zero to a maximum value. The mass will begin to move on the surface once the applied force is greater than μsFN,.
Once the object starts to move, an external force must be applied in order to keep the object moving to overcome fricon due to an object sliding over a surface. This type of fricon is called kinec fricon. The equaon for kinec fricon is Ff = μkFN where μk is the coefficient of kinec fricon. Noce that this is an equaon and not an inequality. As long as a mass is sliding across a surface, the force due to kinec fricon will stay the same. In order for a mass to speed up, the applied force must be greater than the friconal force, otherwise it will slow down. In order to maintain a constant velocity, the applied force must be equal to the force of kinec fricon.
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© 2014 eScience Labs, LLC.All Rights Reserved
Click to Run
Lab 4/Lab4Friction - Copy.docx
Lab 4 Friction PHY250L
Student Name: Click here to enter text.
Kit Code (located on the lid of your lab kit): Click here to enter text.
“Pre-Lab Questions”
2. “Applying Newton’s Second Law and the equation for static friction (F = μsN)₁ prove that the coefficient of static friction (μs) is related to the minimum angle, θ, that causes the block to slip (Figure 5) by the equation μs = tan(θ).” When the angle (theta) is increased, friction is lessened due to lesser normal force, causing the block to slide faster.
3. “A person applied a horizontal force to a crate of mass, m, that caused the crate to move at a constant velocity (Figure 6). Show that the relationship between the applied force and the normal force is FA = μFN.”
Applied force and frictional force are equal. The relationship is true because the scenario states that there is a constant velocity.
“Experiment 1: Static Friction and Mass on an Inclined Plane”
“Table 1: Wooden Block Incline Data” “Table 2: Metal Washer Incline Data”
|
“Trial” |
“Angle” |
|
“1” |
20 |
|
“2” |
23 |
|
“3” |
21 |
|
“4” |
22 |
|
“5” |
20 |
|
“Trial” |
“Angle” |
|
“1” |
30 |
|
“2” |
33 |
|
“3” |
32 |
|
“4” |
30 |
|
“5” |
31 |
“Post-Lab Questions”
1. “Using the result from Pre-Lab Question 2, calculate the coefficient of friction for each angle in Table 1 and Table 2. Find the average value for the coefficient of friction for wood and metal on cardboard.”
Click here to enter text.
2. “Comment on your coefficients of static friction for wood and metal. If they are different, why do you think they are different? If they are the same, why do you think they are the same?”
Click here to enter text.
“Experiment 2: Static Friction vs. Kinetic Friction”
“Table 3: Peak Static Frictional Force”
|
Total Mass (kg) |
Trial 1 (N) |
Trial 2 (N) |
Trial 3 (N) |
Average (N) |
|
.5 |
2.6 |
2.4 |
2.2 |
2.4 |
|
.75 |
3.0 |
2.8 |
2.8 |
2.86 |
|
1 |
3.2 |
3.2 |
3.4 |
3.26 |
“Table 4: Kinetic Frictional Force”
|
Total Mass (kg) |
“Trial 1 (N) |
“Trial 2 (N) |
Trial 3 (N) |
Average (N) |
|
.5 |
1.6 |
1.6 |
1.6 |
1.6 |
|
.75 |
2.2 |
2.2 |
2.2 |
2.2 |
|
1 |
2.8 |
2.8 |
2.8 |
2.8 |
Calculate the average force from Trials 1, 2, and 3 for static and kinetic friction.
Record the averages in Tables 5 and Table 6.
|
Average Applied Force (N) |
Normal Force (N) |
|
Click here to enter text. |
Click here to enter text. |
|
Click here to enter text. |
Click here to enter text. |
|
Click here to enter text. |
Click here to enter text. |
|
Average Applied Force (N) |
Normal Force (N) |
|
Click here to enter text. |
Click here to enter text. |
|
Click here to enter text. |
Click here to enter text. |
|
Click here to enter text. |
Click here to enter text. |
“Table 5: Static Frictional Force”
Table 6: Kinetic Frictional Force
“Post-Lab Questions”
1. “From the result from Pre-Lab Question 3, the relationship between the applied force and the normal force is FA = μFN. When the data for applied force vs normal force is plotted, the slope of the graph is equal to the coefficient of friction. Plot graphs of average force vs normal force from the data in Table 5 and Table 6.”
Please draw a diagram to include on this document.
“Insert photo of diagram with your name clearly visible in the background:””
“Insert photo of experimental setup with your name clearly visible in the background:””
Lab 5/Lab 5- Conservation of Energy Material.pdf
1/4/2017 Lab 5: Conservation of Energy
https://moodle.straighterline.com/pluginfile.php/72232/mod_resource/content/5/CourseRoot/html/lab010s001.html 1/7
Learning Objec�ves
Relate energy to work Calculate the amount work done by a force Compare and contrast types of energy Apply The Law of Conserva�on of Energy to poten�al and kine�c energy
Introduc�on
You may be familiar with the common usage of the word energy. For example, you may grab an energy drink to get through the night when cramming late at night for a test the next morning, If you're an athlete, you may reach for an energy bar to help you finish a game. Energy is one of the central concepts in science; it has been used to explain many natural phenomena. But what is energy, exactly? Energy is defined as the ability to do work. The amount of energy an object has equals the amount of work it has the ability to do. Think of
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Figure 1: Work done on a tray.
energy as the currency for performing work. For example, 100 Joules of energy is required to do 100 Joules of work.
Work Done by a Constant Force
Work is done when a force causes a mass to move a distance. The unit for work is called the Joule (J), which is defined as a force of one Newton ac�ng over one meter. Quan�ta�vely, the defini�on of work is:
Work = F∙d
where F is the force applied, and d is the displacement. Both force and displacement are vector quan��es, and the opera�on in the equa�on above is known as the dot product. It is one of the forms of vector mul�plica�on, which yields a scalar quan�ty. This equa�on for work is used for a constant force along straight line displacement. We can simplify the equa�on above to be:
Work = F||∙d
In this equa�on, F|| is the amount of force applied along or against the displacement, d. We can also write this equa�on as:
Work = Fd cosθ
where F is the magnitude of the force, d is the displacement, and θ is the angle between the direc�on of the force and displacement. The direc�on factor is a cri�cal aspect of work to understand. For example, a waiter exerts a force to hold a serving tray at a steady height (Figure 1). Is work done on the tray? It may be surprising that the answer is no. Although it takes energy to keep the tray raised, and the waiter is moving across the room, the direc�on of the applied force is not the same as the direc�on of mo�on. Therefore, there is no work done on the tray by the force applied to it.
Work Done by a Variable Force
Forces are not always constant and may change with an object’s loca�on. Springs are a good example: as a spring is stretched, it must be pulled harder to keep the spring stretched out. When the force is only varying in the x‐component and has a straight line displacement, mathema�cally, this can be solved using integra�on:
where F(x) is the variable force integrated with respect to posi�on from its original posi�on, x0, to its final posi�on, xf. Graphically, this is equivalent to calcula�ng the area under the curve of the force versus posi�on graph (Figure 2).
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Figure 3: Wind turbines.
Figure 2: The area under the plot equals the amount of work done.
If the force is variable and the displacement is along a curved path, work can be found by using the following equa�on:
where F is a variable force integrated from one point to another with respect to an infinitesimally small displacement vector (dl) that lies tangent to the path of its posi�on. Once again the dot product is present but can be simplified in certain cases where Φ is the angle between F and dl.
Types of Energy
Energy is one of the most important concepts in physics, and comes in a variety of forms ‐ chemical, gravita�onal, elas�c, electric, and rota�onal energy to name a few. Of these many forms, energy can generally be categorized as kine�c or poten�al energy. Kine�c energy (KE) is the energy associated with the state of mo�on of an object. Wind energy is an example of kine�c energy (Figure 3). The molecules of gas in the air are in constant mo�on, providing energy that can be harnessed by wind turbines like the ones above that convert it into electric energy. The faster the object is moving, the greater its KE. The rela�onship between kine�c energy (KE), mass (m) and velocity (v) is:
KE = ½ mv2
In contrast, poten�al energy (PE) is the stored energy associated with the posi�on of an object. This type of energy exists when there is some kind of force that returns an object to its
1/4/2017 Lab 5: Conservation of Energy
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Figure 4: Water stored in the Shasta dam in California for hydroelectricity.
original posi�on a�er being displaced. For instance, an object li�ed above the ground has gravita�onal poten�al energy because it will accelerate to the ground when released. Water stored in a dam for hydroelectricity genera�on is also a form of poten�al energy. When the valves open to allow the water to flow, the gravita�onal poten�al energy of the water is converted to kine�c energy (Figure 4). The more massive the object and the greater the height, the more energy the object has when it falls to the ground. The equa�on to determine gravita�onal poten�al energy (PEgravity) is:
PEgravity = mgh
where m is the mass of the object, g is the gravita�onal force (9.8m/s2), and h is the height above the Earth's surface.
There are �mes when poten�al energy can be found, and the force on the object is needed. By taking the deriva�ve of poten�al energy with respect to posi�on, this gives us the force on the object. In one dimension the equa�on is:
for this equa�on U is poten�al energy measured in Joules. From this equa�on, we can also relate work to poten�al energy:
W = ‐△U
Another example of poten�al energy is a rubber band. A stretched rubber band has the poten�al to return to its original length, doing work in the process (force applied over a distance). In this sense, poten�al energy is o�en referred to as the energy of posi�on, as it is energy that depends on how far an object is removed from a posi�on of equilibrium. Similarly, a spring has elas�c poten�al energy that increases as it is stretched, and also depends on the material it is made from. With increased s�ffness in springs comes the ability to store more energy and do more work.
? Did You Know...
As the cyclist rides through the mountain trail, chemical energy is converted into kine�c energy and thermal energy (you get hot riding a bike). As the cyclist climbs the mountain, the kine�c energy in the bicycle and the cyclist will be converted to other forms of energy. Fric�on between the �res and ground causes kine�c energy to transform to heat energy. If a cyclist approaches a hill with enough speed, no more peddling will be necessary to reach the top because the kine�c energy will be converted to poten�al energy. If the cyclist needs to stop and applies the brakes, kine�c energy is dissipated through fric�on as heat energy.
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Work Energy Theorum
Work is related to the object's displacement as can be seen by the equa�ons discussed earlier. However, work is also dependent upon the change in an object's speed. If an object is speeding up then work is posi�ve, if the object is slowing down then work is nega�ve, and if there is no change in the object's speed then there is no work done. The work‐energy theorem says that:
Work = K2 ‐ K1 = △U
This means that the change in a object's kine�c energy equals the work done by the force on the object. Remember that kine�c energy equals 1/2mv2.
Power
Power is the rate at which work is done, so power is analyzing the work being done over �me. Just as there is average velocity and instantaneous velocity, there is average power and instantaneous power. To find average power, you look at the change in work over a �me period:
For instantaneous power, you take the deriva�ve of work with respect to �me:
The standard unit of power is a wa� (W) where 1 W is equal to 1 J/s. Power can also be found by looking at the force ac�ng upon an object and the object's velocity. In this case, average power can be found by the given equa�on:
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Where F|| is the component of the force that acts parallel to the displacement vector and vaverage is the average velocity. To find the instantaneous power by looking at force and velocity, the dot product can again be used:
Conserva�on of Energy
Energy can change forms, such as poten�al energy transforming into kine�c energy. The law of conserva�on of energy states that the energy of an isolated system cannot be created or destroyed; it can only change forms or be transferred from one object to another. Consider a system composed of a ball and the Earth. If a ball is sta�onary three meters above the ground, it has gravita�on poten�al energy. When the ball falls due to the force of gravity (remember Earth is part of our system), the kine�c energy of the ball right before it hits the ground is equal to the stored gravita�onal poten�al energy of the ball at three meters above the ground. When the ball bounces back up, the kine�c energy will be converted back into poten�al energy. Since the total energy of the system is equal to the original poten�al energy at three meters, the maximum height the ball can ever bounce to is three meters. A ball rarely bounces back to the original height because some energy goes into hea�ng the air and ground through air resistance and contact with the ground.
As long as all energies are accounted for, the sum of the energies is the same at any moment. This is a very useful principle because now the energy of an object at any two moments in �me can be compared! Let's examine this quan�ta�vely using the ball and Earth system.
Problem: A ball of mass, m, is dropped from rest from a height, h, of 3 meters. What is the ball's speed a�er it has fallen 2 meters?
Solu�on: If the ground is chosen as a reference height (where the height equals zero meters), the two states to consider are when the ball is at rest at 3 meters above the ground, and a�er is has fallen 2 meters, or it is located 1 meter above the ground. The total energy of the first state, E1, is equal to the total energy of the second state, E2:
E1 = E2
Energy converted to heat by air fric�on will be ignored. This means that the only energies the ball can have during the drop is kine�c and poten�al energy:
KE1 + PE1 = KE2 + PE2
When the kine�c energy states and poten�al energy states are replaced with their equivalents the following equa�on can be derived:
½ mball v1 2 + mball gh1` = ½ mball v2
2 + mball gh2
No�ce the mass of the ball is in every term and can be divided out of the equa�on. In E1, the ball is dropped from rest which means v1 = 0 m/s.
gh1` = ½ v2 2 + gh2
Lab 5/Lab 5- Conservation of Energy.pdf
1/4/2017 Lab 5: Conservation of Energy
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Experiment 2: Conservaꭍ�on of Energy‐ Data Analysis
In this experiment you will apply your knowledge of Conservaꭍ�on of Energy to a given data set. Table 2 contains data for the posiꭍ�on of a ball with mass 0.5 kg that has been dropped from 5 meters from rest at given ꭍ�me points. In order to test the conservaꭍ�on of energy, the potenꭍ�al energy and kineꭍ�c energy of the ball need to be calculated.
Materials
Table 2
Procedure
1. To calculate the kineꭍ�c energy, the velocity of the ball needs to be known. From the given table, only posiꭍ�on is provided. To find velocity, a method called the "leap‐frog” method can be used to approximate the velocity using the posiꭍ�on and ꭍ�me points just before and a⚽er:
v= Δx Δt
The first velocity point has been done for you as an example:
v = Δx
= x3 ‐ x1 =
0.10 m‐ 0.00 m = ‐ 0.40 m/s
Δt t3 ‐ t1 0.496 s ‐ 5.00 s
2. Use this method to calculate the other velociꭍ�es. Record the velociꭍ�es in Table 5.
Table 2: Dropped Ball Data
Time (s) Ball Posiꭍ�on
(m) Ball Velocity
(m/s) Potenꭍ�al Energy
(J) Kineꭍ�c Energy
(J) Total Energy
(J)
0.00 5.00 0.05 4.99 0.10 4.96 0.15 4.89
1/4/2017 Lab 5: Conservation of Energy
https://moodle.straighterline.com/pluginfile.php/72235/mod_resource/content/6/CourseRoot/html/lab010s004.html 2/2
0.20 4.78 0.25 4.69 0.30 4.54 0.35 4.40 0.40 4.22 0.45 4.00 0.50 3.80 0.55 3.50 0.60 3.26 0.65 2.93 0.70 2.60 0.75 2.23 0.80 1.88 0.85 1.46 0.90 1.05 0.95 0.58 1.00 0.11
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Lab 5/Lab5ConservationofEnergy.docx - Copy.docx
Lab 5 Conservation of Energy PHY250L
Student Name: Click here to enter text.
Kit Code (located on the lid of your lab kit): Click here to enter text.
“Pre-Lab Questions”
1. “The force due to a spring is variable and is quantified by F = kx, where k is the spring constant and x is the displacement. Given the graph of the force versus displacement graph for a spring in Figure 5, write an equation for the amount of work done by the spring:”
a. “The work done by the spring integration” Click here to enter text.
b. “The work done by the spring by calculating the area under the curve.”
Click here to enter text.
2. “Consider the ball example in the introduction when a ball is dropped from 3 meters. After the ball bounces, it raises to a height of 2 meters. The mass of the ball is 0.5 kg.”
a. “What is the speed of the ball right before the bounce?” Click here to enter text.
b. “How much energy was converted into heat after the ball bounced off the ground? (Hint: Thermal Energy (TE) will now need to be included in your conservation of energy equation and you will now need to know the mass of the ball).”
Click here to enter text.
c. “What is the speed of the ball immediately after the ball bounces off the ground?”
Click here to enter text.
3. “A 2 kg book is sitting on a horizontal glass table top. A woman pushes the book across the table with a horizontal force F. The book’s position changes as a function of time given by x(t) = (2.9 m/s³) t³- (3.6 m/s²) t². What is the book’s velocity at t=1.2 seconds? Calculate the magnitude of the force at t=1.2 seconds. Find the work done by the force during the first 1.2 seconds of its motion.”
Click here to enter text.
4. “Propeller-driven airplanes have engines that develop a thrust (a forward force on the airplane) of 15,000 N. What is the instantaneous power produced by the aircraft when it is flying at 166.67 m/s?”
Click here to enter text.
“Experiment 1: Work Done by a Spring”
“Table 1: Spring Scale Force Data”
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“Force (N)” |
Distance, x (m) |
ForceAverage (N) |
Δ Distance, Δx (m) |
Work (J) |
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“0”0 |
“0” |
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||
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|
Click here to enter text. |
“0.01” |
Click here to enter text. |
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1 |
“0.01” |
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Click here to enter text. |
“0.01” |
Click here to enter text. |
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2 |
“0.02” |
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Click here to enter text. |
“0.01” |
Click here to enter text. |
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|
3 |
“0.03” |
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||
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Click here to enter text. |
“0.01” |
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4 |
“0.04” |
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Click here to enter text. |
“0.01” |
Click here to enter text. |
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5 |
“0.05” |
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“*Note, you will finish completing Table 1 in the Post-Lab Questions section.”
“Post-Lab Questions”
1. “Create a Force vs. Displacement (stretch) graph.”
“Insert photo of graph with your name clearly visible in the background:”
2. “Using the result of Pre-Lab Question 1, calculate the work done by the spring.” Click here to enter text.
3. “The work done by the spring can be broken down by the work done by each 1 cm stretch. Fill in the rest of Table 1 to calculate the average force applied by the spring over each 1 cm stretch.”
4. “Calculate the work done in each segment and determine the total work done by adding all of the segments together. How does this compare to the work done by the spring calculated in Post-Lab Question 2?” Click here to enter text.
5. “A student finds a spring that does not obey Hooke’s Law and stretches it 5.0 cm. The force that he had to use in order to stretch the spring is modeled by Fx= 5.4 N + (-2.3 N/m²) x² + (4.3 N/m³) x³. How much work was required for this task?” Click here to enter text.
“Insert photo of your experimental setup with your name clearly visible in the background:”
“Experiment 2: Conservation of Energy – Data Analysis”
“Table 2: Dropped Ball Data”
|
“Time (s)” |
Ball Position (m) |
Ball Velocity (m/s) |
Potential Energy (J) |
Kinetic Energy (J) |
Total Energy (J) |
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“0.00” |
“5.00” |
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“0.05” |
“4.99” |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
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“0.10” |
“4.96” |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
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“0.15” |
“4.89” |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
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“0.20” |
“4.78” |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
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“0.25” |
“4.69” |
Click here to enter text. |
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Click here to enter text. |
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“0.30” |
“4.54” |
Click here to enter text. |
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“0.35” |
“4.40” |
Click here to enter text. |
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Click here to enter text. |
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“0.40” |
“4.22” |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
Click here to enter text. |
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“0.45” |
“4.00” |
Click here to enter text. |
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“0.50” |
“3.80” |
Click here to enter text. |
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“0.55” |
“3.50” |
Click here to enter text. |
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“0.60” |
“3.26” |
Click here to enter text. |
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“0.65” |
“2.93” |
Click here to enter text. |
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“0.70” |
“2.60” |
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“0.75” |
“2.23” |
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“0.80” |
“1.88” |
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“0.85” |
“1.46” |
Click here to enter text. |
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Click here to enter text. |
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“0.90” |
“1.05” |
Click here to enter text. |
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Click here to enter text. |
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“0.95” |
“0.58” |
Click here to enter text. |
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Click here to enter text. |
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“1.00” |
“0.11” |
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“Post-Lab Questions”
1. “Graph the potential energy, kinetic energy, and total energy of the ball.”
“Insert photo of graph with your name clearly visible in the background:”
2. “Describe the shape of each graph.” Click here to enter text.
3. “What are the limitations of using the leap-frog method?” Click here to enter text.
4. “An object is moving horizontally with no vertical movement, and it has a mass of 0.20 kg. Its potential energy function can be described as U(x) = (3.8 J/m²) x² — (2.1 J/m) x. What is the force on the object? What is the magnitude of the acceleration at x = 0.44 meters?” Click here to enter text.
