Part 1 A 17.1 kg block is dragged over a rough, horizontal surface by a constant

Question

Part 1

A 17.1 kg block is dragged over a rough, horizontal surface by a constant force of 143 N
acting at an angle of angle 29
?
above the horizontal. The block is displaced 91.7 m and the
coe?cient of kinetic friction is 0.163.Find the work done by the 143 N force. The
acceleration of gravity is 9.8 m/s
2
.
Answer in units of J
Part2

Find the magnitude of the work done by the
force of friction.
Answer in units of J

Part 3

Find the work done by the normal force.
Answer in units of J

Part 4

What is the net work done on the block?
Answer in units of J

Explanation / Answer

The block is moved in the x direction so to find the work done by the 143N force we have to find the part of the force in that direction.

cos(29)=x/143

x=125.07N

Work done by the 143N force=125.07N*91.7m= 11468.98J   ans

part 2: the friction force is equal to the normal force times the friction coefficient

normal force=force due to gravity-y direction of the 143N force

y direction of the 143N force=sin(29)*143

=69.32N

force due to gravity=17.1kg*9.81m/s2

=167.751N

normal force=167.751-69.32

=98.42N

force due to friction=98.42*0.163=16.043N

work done by the friction force=16.043N*91.7m=1471.14J        ans

part 3: work done by the normal force is the force times the displacement.

The normal force acts in the upward, y direction and there is no displacement in that direction.

since there is no displacement, the work done is 0.

part 4: there is only displacement in the x direction so we will find work in that direction

net work=work done by the 143N force in x direction - work done by the friction force

net work=11468.98J-1471.14J= 9997.83J

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