Question

A factory worker moves a 21.0 kg crate a distance of 4.80 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.300.

What magnitude of force must the worker apply? (N)

How much work is done on the crate by the worker's push? (J)

How much work is done on the crate by friction? (J)

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

How much work is done by gravity? (J)

What is the net work done on the crate? (J)

Answer #1

Part A

Using force balance

Fnet = Fw + Ff = 0

Fw = -Ff

Fw = -(-uk*m*g)

Fw = force applied by worker = 0.30*21*9.8 = 61.74 N

Part B

Workdone by worker

W = F.d = F*d*cos theta

theta = angle between force and distance = 0 deg

Ww = 61.74*4.80*cos 0 deg = 296.36 J

Part C.

Wf = Ff*d*cos theta

for friction force, theta = 180 deg

Wf = 61.74*4.80*cos 180 deg = -296.35 J

Part D

theta = angle between normal force and distance = 90 deg

Cos 90 deg = 0

Wn = Fn*cos 90 deg = 0 J

Part E

theta = angle between gravity and distance = 90 deg

Cos 90 deg = 0

Wn = Fg*cos 90 deg = 0 J

Part F

Wnet = Ww + Wf + Wn + Wg

Wnet = 296.35 + (-296.35) + 0 + 0 = 0 J

Please Upvote.

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