Question is below b/c its length does not fit into the summary (please answer asap)

Question

A 6.00 kg block slides down a 3.40 m long inclined plane. The speed of the block at the bottom is 5.40 m/s. If the height of the plane is 2.00 m, what is the work done by friction?

 

Im confused on how to write the net force equation. I know that theta= 36 degrees. 

 

Answer 1

You don't need a force equation. You know that the block is raised 2m off the ground initially. So its gravitational energy is

mgh=6•9.81•2=117.72 J.

 

At the bottom of the ramp, it has no gravitational energy, but its kinetic energy is 0.5mv²=0.5•6•(5.40)²=87.48 J. So in sliding down the ramp, you lost 117.72-87.48=30.24 J to friction.