Question

consider a simple situation: a 12.9kg crate slides rightward from rest without friction 3.2m down a ramp inclined 14.6

Answer 1

A factory worker pushes a 30.3 kg crate a distance of 4.8 m along a level floor at constant velocity by pushing downward at an angle of 28 below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.

a) What magnitude of force must be applied to move the crate at constant velocity?
Total normal force = (mg + Fsin 28)
Friction = ? (mg + Fsin 28)
constant velocity implies the applied force F cos 28 is equal to the frictional force.
? (mg + F sin 28) = F cos 28
0.25. (30.3*9.8 +0.47F) = 0.88F
F= 97.36 N

b) How much work is done on the crate by this force when the crate is pushed a distance of 4.8 m?
W= 97.36 * cos 28* 4.8 = 412.63 J

c) How much work is done on the crate by friction during this displacement?
Since 412.63 J is the work done against friction
Wf= 412.63 J

d) How much work is done by the normal force?
Wnf= N*0 J

e) How much work is done by gravity?
Wg=No displacement in the direction of Wg and hence the work done is zero.
f) What is the total work done on the crate?
Wnet = 412.63 J