physics: work due to friction???

Question

A 56.00 kg keg of beer slides upright down a 2.75 m long plank leading from the back of a truck 1.16 m high to the ground. The coefficient of kinetic friction, muk= 0.10.

a) Determine the amount of work done on the keg by gravity.

Fgrav = ma = 56(9.8)
Wgrav = ma*d = 56(9.8)(1.16) = 636.6 J

b) What is the total work done by friction on the keg?

I'm not sure how to go about doing the second part. I tried 548.8 N * 2.75 m = -150.92 J, but it didn't work. In this situation, is the work due to gravity positive, or am I wrong in my equation?

Answer 1

For part A you are right.
Now, for part b, you can calculate the force against the plank.

You can calculate the angle of the plank as
sin(theta)=1.16/2.75

The weight of the keg has two components:
parallel to the plank, and perpendicular to the plank

The perpendicular component
cos(theta) times the coefficient of friction times the displacement is the work:

cos(theta)*56*g*.1*2.75
=136 j

j