A factory worker pushes a 28.0 kg crate a distance of 3.5 m along a level floor
ID: 1658861 • Letter: A
Question
A factory worker pushes a 28.0 kg crate a distance of 3.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.25.
a) What magnitude of force must the worker apply?
b) How much work is done on the crate by this force?
c) How much work is done on the crate by friction?
d) How much work is done on the crate by the normal force?
e) How much work is done on the crate by gravity?
f) What is the total work done on the crate?
Explanation / Answer
a)
Fr = u*Fnormal
Fr = u*mg
Fr = 0.25 x 28 x 9.81
Fr = 68.67 N
F = Fr = 68.67 N
b)
W = F x d x cos(theta)
W = 68.67 x 3.5 x cos(0)
W = 240.345 J
c)
W = Fr x d x cos(theta)
W = 68.67 x 3.5 x cos(180)
W = -240.345 J
d)
Normal forces perform no work, gravity perform no work since their directions are 90 degrees to the velocity.
e)
Total work done on the crate = total force x distance = 0 x 3.5 m = 0 joule
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