A factory worker pushes a 30 kg crate a distance of 4.5 meters along a level flo

Question

A factory worker pushes a 30 kg crate a distance of 4.5 meters along a level floor at constant velocity by pushing it downward at an angle of 30 degrees below the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.25.

A. What magnitude of force must the worker apply to move the crate at constant velocity?
B. How much work is done on the crate by this force when the crate is pushed a distance of 4.5 meters?
C. How much work is done on the crate by the normal force? By gravity?
D. What is the total work on the crate?

Explanation / Answer

a) force of friction = R where R is reaction force on crate from ground. Since the person pushes the crate down, the total reaction force R = mg + force applied by worker in verticle direction.

Thus, R = (30*9.8) + Fsin30 = 294 + 0.5F where F is force applied by worker.

Thus, friction f = 0.25 [ 294 + 0.5F] = 73.5 + 0.125F.

Force applied by worker in horizontal direction = Fcos30

For constant velocity, Fcos30 = f.

Thus, 0.866F = 73.5 + 0.125F.

Thus, F = 99.19N.

b) Work done = force x distance moved in direction of force.

crate moves only in horizontal direction so work done =horizontal force x horiontal distance = 4.5*Fcos30 = 386.555J.

c) 0. because work done = force * distance moved IN DIRECTION OF FORCE.

d) work done by person = 386.555J. Work done by other forces = 0.

so total work done = 386.555J

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