A factory worker pushes a 29.2kg crate a distance of 4.1m along a level floor at constant velocity by pushing downward at an angle of 28 degrees below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.

Question

(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 frication?
(d) How much work is done on the crate by normal force?
(e) How much work is done on the crate by the gravity?
(f) What is the total work done on the crate?

Answer 1

weight = m g
normal force down on floor N= m g+Fsin 28
friction force back = .25 N
so
F cos 28 = friction force back

(a) F = friction force back / cos 28

force component in direction of motion = F cos 28 Newtons
d = 4.1 meters
so
(b) Win = 4.1 F cos 28 Joules
(c) -4.1 F cos 28
(d) no vertical motion so = ZERO
(e) same as (d) ZERO
(f) zero, there is no net force on the crate so the work in from the push is equal and opposite to that out by friction.