As shown in the figure below, a box of mass m = 58.0 kg (initially at rest) is p
ID: 1537492 • Letter: A
Question
As shown in the figure below, a box of mass
m = 58.0 kg
(initially at rest) is pushed a distance
d = 89.0 m
across a rough warehouse floor by an applied force of
FA = 230 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 = __________ J
(c) work done by the normal force
WN = __________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
Please answer all parts!!!
30 rough surface iExplanation / Answer
Make the force components.
Fx = F*cos30 = 230*cos30 = 199.185 N
Normal force = m*g + F*sin30
N = 58*9.81 + 230*sin30 = 683.98 N
Friction force Fr = u*N
Fr = 683.98*0.1 = 68.398 N
(a)
Work done by applied force = Fx*d = 199.185*89 = 17727.54 J
(b)
Work done by force of gravity = 0
(no dispalcement in vertical direction.)
(c)
Work done by normal force = 0
(no displacement in vertical direction.)
(d)
Work done by friction force = - 68.398*89 = -6087.42 J
(e)
Net work = 17727.54 - 6087.42 = 11640.118 J
(f)
Net force = 199.185 - 68.398 = 130.78 N
Work done by net force = 130.78*89 = 11640 J
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