Work Done by Friction Force: A 4.0 kg Block

[IKonquer]

A 4.0 kg block is dragged over a rough horizontal surface by a constant force of 20 N. The speed of the block increases from 3.0 m/s to 5.0 m/s in a displacement of 5.0 m. What is
the magnitude of the work done by the friction force during this displacement?

My work was the following:

Ei + W = Ef

(.5)(m)(3)^2 + W = (.5)(m)(5)^2

And as a result, I got W = 50 - 18 = 32 J.

I'm not sure why this answer is wrong. Thanks in advance.

 

[gneill]

You've calculated the energy that went into raising the speed from 3 to 5 m/s. But you haven't calculated the energy that was lost to friction.

Hint: How much work did the applied force do?

 

[IKonquer]

I see. I just didn't realize that I needed to account for the applied force.

Thanks again.