As shown in he gu re below a box of rmass m = 62.0 kg initially at rest is pushe
ID: 1420447 • Letter: A
Question
As shown in he gu re below a box of rmass m = 62.0 kg initially at rest is pushed a distance d-58.0 m across a rough warehouse floor by an applied force of A 248 N re teda arange 30 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.) rough surface (a) work done by the applied force WA-12456.91 , (b) work done by the force of gravity (c) work done by the normal force (d) work done by the force of friction wf-4243.28 , (e) Calculate the net work on the box by finding the sum of all the works done by each individual force. WNet 821363 (f) Now find the net work by first finding the net force on the box, then finding the work done by this net force. A good diagram showing all the forces and their components will help you determine the correct expression for the net force. Knowing the net force acting on the box and the distance this force WNet acts, how can we determine the net work done on the box? Pay careful attention to the angle between the direction of the net force acting on the box and the displacement vector. JExplanation / Answer
distance moved in hori. direction = 58 m
Fwd force in hori. direction = 248 cos 30 = 214.77 N
a) work done by applied force = 214.77 * 58
= 12456 J
B) work done by gravity = 0
because gravity is acting in vertical direction and no movement of block is there in vertical direction
c) normal force is also acting in vertical direction and no movement of block is there in vertical direction
for vertical direction, toal downward force = toal upward force
so Fsin 30 + mg = N
N = 731.6 N
D) work done by friction = uK N * 58
= - 4243.2 J ( negative because friction is bwd. and movement is fwd. )
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