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(adsbygoogle = window.adsbygoogle || []).push({});

4.84×102 J

b) the increase in internal energy of the box + floor system due to friction 4.67×102 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

I have everything else figured out here I'm stuck. The change of kinetic energy I used the equation ΔK= -fk(d) which would equal -4.66e2 J but I'm not sure if there's another force that I need to add on. If there is where?

f) the final speed of the box: for this I orignally used W=1/2mvf^2- 1/2mvi^2 making it 484=1/2(41)vf^2 which would give a final velocity of 4.86m/s is this right?

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# Homework Help: Change in Kinetic energy and final speed

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