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)
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
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