A crate of 44.8- tools rests on a horizontal floor. You exert a gradually increa

Question

A crate of 44.8- 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 337 . After that you must reduce your push to 201 to keep it moving at a steady 30.0 .
What is the coefficient of static friction between the crate and the floor?

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

What push must you exert to give it an acceleration of 1.28 ?

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 . What magnitude push would cause it to move?

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 44.8x9.8 = 439.04N The maximum static friction is 337N So the coefficient of static friction is 337/(439.04) = 0.7676 b) As the crate is moving at a constant speed, the forces on the crate are balanced. So the kinetic friction is 201 N. The coefficient of kinetic friction is 201 / 439.04 = 0.4578 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 - 201 Applying Newton's second law gives P - 201 = 44.8x1.28 So P = 258.344N d) On the Moon the mass of the crate does not change. But the weight of the crate changes to 44.8x1.62 = 72.576 N . The normal force is also 72.576 N. So the maximum static friction is 0.7676 * 72.576 = 55.7093 N e) The kinetic friction is 0.4578 * 72.576 = 33.2253 N If the magnitude of the push in c) is maintained here, The nett force is 201 - 33.2253 = 167.7747 N Applying Newton's second law gives 167.7747 = 44.8 * a So a = 3.74497 (m/s^2)

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