A factory worker pushes a 28.0 kg crate a distance of 5.4 m along a level floor
ID: 1598978 • Letter: A
Question
A factory worker pushes a 28.0 kg crate a distance of 5.4 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?
N
(b) How much work is done on the crate by this force?
J
(c) How much work is done on the crate by friction?
J
(d) How much work is done on the crate by the normal force?
J
(e) How much work is done on the crate by gravity?
J
(f) What is the total work done on the crate?
J
Explanation / Answer
(A) in vertical,
N - m g = 0
N = 28 x 9.8 = 274.4 N
f = uk N = 68.6 N
in horizontal, Fnet = m a = 0
F - f= 0
F = f = 68.6 N
(B) Work done = F.d = F d cos(theta)
by force = 68.6 x 5.4 x cos(0) = 370.44J
(C) by friction = 68.6 x 5.4 x cos180 = - 370.44 J
(D) by normal = 274.4 x 5.4 x cos90 = 0
(E) by gravity = 274.4 x 5.4 x cos90 = 0
(f) total work done = summation of work done by all forces = 0
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.