As shown in the figure below, a box of mass m = 64.0 kg (initially at rest) is p
ID: 1551603 • Letter: A
Question
As shown in the figure below, a box of mass m = 64.0 kg (initially at rest) is pushed a distance d = 64.0 m across a rough warehouse floor by an applied force of FA = 222 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 = J (b) work done by the force of gravity Wg = Correct: Your answer is correct. J (c) work done by the normal force WN = 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 = 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 = J
Explanation / Answer
F _p= FA cos theta = 222 cos30 = 192.25 N
F_N = mg + F sin theta = 64 (9.8) + 222 sin 30 = 738.2 N
fk = u F_N = 0.1 ( 738.2) = 73.82 N
(a)
W_F= 192.25 ( 64) = 12304 J
(b)W_g = 0 ( since nor vertical diplacment)
(c) W_N = 0 ( since nor vertical diplacment)
(d)
W_f = fk d = 73.82 ( 64) = -4724.48 J
(e) W_ net = W_F-W_f =7579.52 J
(f)
net force = 192.25 N-73.82 N=118.43 N
the work done by this net force
W= 118.43 N(64) = 7579.52 J
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