Setup of the Quantitative Description of Your Rube Goldberg Device Step

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Running head: QUALITATIVE DESCRIPTION OF YOUR RUBE GOLDBERG DEVICE STEP 1

QUALITATIVE DESCRIPTION OF YOUR RUBE GOLDBERG DEVICE STEP 2

Qualitative Description of Your Rube Goldberg Device Step

Calvin Singh

SNHU

PHY 101: Milestone One.

Qualitative Description of Your Rube Goldberg Device Step.

I. Step Selection

a. Select a step or stage in the Rube Goldberg device. Provide a concise description of the step.

In this step, I will have a rubber block weighing 5kg on a table being pulled by a falling weight connected via a pulley using a chain. The rubber block rests 10cm away from the edge of the table and attains an acceleration of 3m/s^2 in 2 seconds before falling off the table’s edge. The static friction coefficient (µs) is 1.0, while the kinetic friction coefficient (µk) is 0.7.

II. Selected Step A. Initial Velocity

a. Calculate the initial velocity of the object in the selected step.

The initial velocity of the rubber block is 0m/s. This is because the rubber block is resting stationary on the table. According to Newton’s first law of motion, a body at rest will stay at rest until a net external force acts on it (Suleiman, 2018). In the setup, the net external force due to the falling weight that will overcome the friction force between the rubber block and the surface of the table

b. What does the initial velocity of the object tell you about the behavior of the object?

The 0m/s velocity implies that the rubber block remains in a state of inertia until the net force due to the falling weight causes it to move. When the falling weight pulls the rubber block, the rubber block simultaneously pulls the falling weight. These forces are opposite in direction and equal in size.

Selected step B. Velocity and Force Calculations

c. Calculate the change in velocity that would be observed based on kinematics and force principles.

Change in velocity = Final velocity – Initial velocity.

Final velocity = displacement / time.

Final velocity = 10 cm / 2 seconds = 5cm/s =0.05m/s.

Change in velocity = 0.05m/s.

d. Then, use Newton’s Second Law to calculate the force acting on the object.

Newton’s second states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object (Roeder, 2017).

F = ma.

In the set-up, the force that causes the rubber block to accelerate towards the table’s edge, ma is equal to the force due to the weight of the falling weight, mg. These forces are equal in size and exert in the opposite direction.

F = ma = mg = 5kg × 3m/s^2 = 15N.

Considering the effects of static and kinetic friction,

Static friction, Ff max = µsF =1.0 × 15 = 15N.

Kinetic friction, Ff = µkF =0.7 × 15 = 10.5N

References.

Roeder, J. L. (2017). More on deriving Newton’s second law. The Physics Teacher55(7), 388-388.

Suleiman, R. (2018). Newton's First Law revisited. Journal of Modern Physics.