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A man pulls a crate of mass 9 kg which is on frictionless bearings up a ramp that is h = 5 m high and angled at 37°. The crate goes from rest at the bottom to rest at the top of the incline. (Hint: Remember the direction of force and displacement.)

(a) How much work is done by the man?

(b) How much work is done by gravity?

(c) How much work is done by the normal force?

1 Answer

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Step-by-step explanation:

To solve this problem, we need to use the work-energy theorem, which states that the work done on an object equals the change in its kinetic energy.

(a) The work done by the man is equal to the change in kinetic energy of the crate. Since the crate starts from rest at the bottom and comes to rest at the top, its initial and final kinetic energies are both zero. Therefore, the work done by the man is equal to the potential energy gained by the crate as it moves up the ramp:

W_man = mgh = (9 kg)(9.81 m/s^2)(5 m) = 441.45 J

where m is the mass of the crate, g is the acceleration due to gravity, h is the height of the ramp, and W_man is the work done by the man.

(b) Gravity does negative work on the crate because it acts in the opposite direction to the displacement of the crate. The work done by gravity is equal to the negative change in potential energy of the crate:

W_gravity = -mgh = -(9 kg)(9.81 m/s^2)(5 m) = -441.45 J

where W_gravity is the work done by gravity.

(c) The normal force acts perpendicular to the displacement of the crate and therefore does no work. Therefore, the work done by the normal force is zero:

W_normal = 0 J.

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