Question

Help!

A frictional force of .2N acts on a .4kg air cart for a distance of .6m. What is the cart's final kinetic energy if it initially had a kinetic energy of .5J?

Answer 1

The work done by anything is W=F*d*cos(theta) where F is the force you're talking about, d is the distance the force is acting through, and theta is the angle between F and D. So in this case, the work done by friction is:
W = .2*.6*cos(180) = -.12 Joules

So friction ate up .12 of our joules! That's ok, we started with .5 Joules so we still have .5-.12=.38 Joules left

Answer 2

Thank you! :)