A crate of 45.2-kg tools rests on a horizontal floor. You exert a gradually incr

Question

A crate of 45.2-kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that the crate just begins to move when your force exceeds 310N . After that you must reduce your push to 231N to keep it moving at a steady 25.2cm/s .

Part A: What is the coefficient of static friction between the crate and the floor?

Part B: What is the coefficient of kinetic friction between the crate and the floor?

Part C: What push must you exert to give it an acceleration of 1.13m/s^2 ? (in N)

Part D: Suppose you were performing the same experiment on this crate but were doing it on the moon instead, where the acceleration due to gravity is 1.62 m/s^2. What magnitude push would cause it to move? (in N)

Part E: What would its acceleration be if you maintained the push in part C?

Explanation / Answer

To cause the crate starting moving the force you exert on it must exceeds the maximum static friction, which can be found by
maximum static friction=(coefficient of static friction)x(normal force).
After it starts moving, the friction exerting on the crate becomes smaller. we call it the kinetic friction. So after it started we need a smaller force to maintain its movement.

a)
The normal force is 45.2x9.8 = 442.96N
The maximum static friction is 310N
So the coefficient of static friction is 310/(442.96) = 0.6998374571067365

b)
As the crate is moving at a constant speed, the forces on the crate are balanced. So the kinetic friction is 231 N.
The coefficient of kinetic friction is 231 / 442.96 = 0.5214917825537295

c)
The nett force acting on the moving crate is the difference between the push and the kinetic friction. Suppose the push is P, then the nett force is = P - 231

Applying Newton's second law gives
P - 231 = 45.2x1.13
So P =282.076 N

d)
On the Moon the mass of the crate does not change. But the weight of the crate changes to 45.2x1.62 = 73.224 N . The normal force is also 73.224 N. So the maximum static friction is 0.699 * 73.224 =51.183576N

e)
The kinetic friction is 0.521 *73.224 = 38.149704 N
If the magnitude of the push in c) is maintained here, The nett force is 231 - 38.149 = 192.850296 N
Applying Newton's second law gives

192.850296 = 45.2 * a
So a =4.266599469026549 (m/s^2)

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