A 40.5 kg box initially at rest is pushed 4.40 m along a rough, horizontal floor with a constant applied horizontal force of 125 N. If the coefficient of friction between box and floor is 0.300, find the following.(d) the work done by the gravitational force J(e) the change in kinetic energy of the box J(f) the final speed of the box m/s

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Asked Nov 22, 2019
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A 40.5 kg box initially at rest is pushed 4.40 m along a rough, horizontal floor with a constant applied horizontal force of 125 N. If the coefficient of friction between box and floor is 0.300, find the following.

(d) the work done by the gravitational force
 J

(e) the change in kinetic energy of the box
 J

(f) the final speed of the box
 m/s

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Expert Answer

Step 1

(d). the work done by the gravitational force is zero. Because the gravitational force on the box and displacement of the box are perpendicular. So the gravitational force on the box is pointed downward.

= Fxcos 90
=0
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= Fxcos 90 =0

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Step 2

(e).
The net work done is equal to the sum of work done by the applied force, work done by the frictional force, work done by the normal force, and work done by the gravitational force.

W
friction
W W
applied F
et
= F.
applied
x+Ffiction * x)+ W. + W,
= F
x+(-umgx)+ 0 +0
applied
=(125)(4.40)-(0.300) (40.5) (9.8) (4.40) +0 +0
=26.1J
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W friction W W applied F et = F. applied x+Ffiction * x)+ W. + W, = F x+(-umgx)+ 0 +0 applied =(125)(4.40)-(0.300) (40.5) (9.8) (4.40) +0 +0 =26.1J

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Step 3

According to the work energy theorem, the net work done is...

ΔΚΕ -W.
net
- 26.1J
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ΔΚΕ -W. net - 26.1J

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