Change in Kinetic energy and final speed

[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
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?

 

[andrevdh]

b) 0 J - The floor experiences a similar friction force, but in the opposite direction. Although we know that both will heat up so this one I am not sure about.

e) The change in kinetic energy of the box is given by the work done by the resultant force acting on the box.

f) use the previous answer

 

[bearhug]

For e) can you explain a little more than that? I'm not sure if I'm going in the right direction in solving that problem.

 

[andrevdh]

Lets come back to b) first. The work done by friction changes the internal energy of a system (the object heats up and therefore stores energy that could under certain circumstances do work). So the change in internal energy of a system due to a frictional force f acting on it over a distance d is given by

\Delta I = -fd

This means that the internal energy change is positive due to a frictional force acing on the object. In this case the increase in the internal energy of the floor and the box comes from a decrease in in internal energy of whoever is pushing the box along the floor.