A factory worker pushes a 28.6 kg crate a distance of 4.4 m along a level floor

Question

A factory worker pushes a 28.6 kg crate a distance of 4.4 m along a level floor at constant velocity by pushing downward at an angle of 32 below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.26.

What magnitude of force must the worker apply to move the crate at constant velocity?

How much work is done on the crate by this force when the crate is pushed a distance of 4.4 m ?

How much work is done on the crate by friction during this displacement?

How much work is done by the normal force?

How much work is done by gravity?

What is the total work done on the crate?

Explanation / Answer

(a)

Total normal force = (mg + Fsin32)

Friction = (mg + Fsin32)

constant velocity implies the applied force F cos 28 is equal to the frictional force.

(mg + F sin32) = F cos32

0.26 * (28.6 * 9.81 + 0.5299 * F) = 0.848 * F

F= 102.7 N

Magnitude of force applied by the worker is 102.7 N

(b)

Work done W = 102.7 * cos 32 * 4.4 = 383.23 J

Work done on the crate by the force = 383.23 J

(c)

Since 383.23 J is the work done against friction

Work done by the friction Wf = -383.23 J

(d)

As, normal force is perpendicular to the direction of displacement,

Work done by the normal force = 0 J

(e)

Similarly, as the gravity is perpendicular to the direction of displacement,

Work done by the gravity force = 0 J

(f)

Total work done on the crate = 383.23 J

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