# Change in kinetic energy of a box (1 Viewer)

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#### bearhug

A 41 kg box initially at rest is pushed 4 m along a rough, horizontal floor with a constant applied force of 121 N. If the coefficient of kinetic friction between the box and the floor is 0.29, find
(a) the work done by the applied force W= FΔx
Ans: 4.84e2 J
b) the increase in internal energy of the box + floor system due to friction ΔEint= fkd fk= ukn= 116.5N
ΔEint= 4.66e2 J
c)the work done by the normal force on the block due to the floor
=0.00 J
d) the work done by the gravitation force on the block due to the earth =0.00 J
e)the change in kinetic energy of the box
ΔK = -fkd + ∑W This is where I'm stuck. Is ∑W= to 4.84e2J because I thought that the applied force was included in the change in kinetic energy but this according to my homework is wrong. Can someone explain what I'm missing here?

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