A factory worker pushes a crate of mass 31.3 kg a distance of 4.40 m along a lev
ID: 1434578 • Letter: A
Question
A factory worker pushes a crate of mass 31.3 kg a distance of 4.40 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 by the normal force?
E) How much work is done by gravity?
F) What is the total work done on the crate?
Please answer all parts, I have tried this multiple times and my math is just not ending up right. Thanks!
Explanation / Answer
let F is the applied force and fk is the kinetic friction.
A) As the crate is moving with constant velocity, the net force acting on it must be zero.
Apply, Fnetx = 0
F - fk = 0
F = fk
= N*mue_k
= m*g*mue_k
= 31.3*9.8*0.25
= 76.7 N
B) Workdone by applied force, W = F*d*cos(0)
= 76.7*4.4*1
= 337.4 J
C) workdone by friction, W = fk*d*cos(180)
= N*mue_k*d*(-1)
= 31.3*9.8*0.25*4.4*(-1)
= -337.4 J
D) Workdone by normal force, W = N*d*cos(90)
= 0
E) Workdone gravity = Fg*d*cos(90)
= 0
F) Total WOrkdone = W_applied force + W_friction + W_normal force + W_gravity
= 337.4 - 337.4 + 0 + 0
= 0
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