As shown in the figure below, a box of mass m = 55.0 kg (initially at rest) is p
ID: 1443956 • Letter: A
Question
As shown in the figure below, a box of mass m = 55.0 kg (initially at rest) is pushed a distance d = 61.0 m across a rough warehouse floor by an applied force of FA = 210 N directed at an angle of 30.0° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the following. (For parts (a) through (d), give your answer to the nearest multiple of 10.)
(a) work done by the applied force WA = 11095.9 Correct: Your answer is correct. J
(b) work done by the force of gravity Wg = 0 Correct: Your answer is correct. J
(c) work done by the normal force WN = 0 Correct: Your answer is correct. J
(d) work done by the force of friction Wf = ????? J
(e) Calculate the net work on the box by finding the sum of all the works done by each individual force. WNet = 7167.5 Correct: Your answer is correct. J
(f) Now find the net work by first finding the net force on the box, then finding the work done by this net force. WNet = ????
Explanation / Answer
a)
applied horizontal force = 210*cos(30) = 181.86 N
work done by applied force = 181.86*61 = 11094 J
b)
force by gravity on box = mg
this force is vertical and
work done = F*d*cos(theta)
here theta = 90 so work done = mg*d*cos(90) = 0 J
c)
Normal force on box = mg
this force is vertical and
work done = F*d*cos(theta)
here theta = 90 so work done = mg*d*cos(90) = 0 J
d)
friction force = coefficient of kinetic friction*Normal foce
friction force = 0.1*55*9.8 = 53.9 N
work done by friction force = 53.9*61*cos(180) = - 3287.9 J
e)
Net work = 11094 + 0 + 0 - 3287.9 = 7806.1 J
f)
net force = horzontal applied force - friction force = 181.86 - 53.9 = 127.96 N
Net work done = 127.96*61 = 7805.56 J
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