The diagram shows a 5kg block of lead released from rest at the top of an incline. The block has a speed of 6 m/s when it reaches the bottom. The angle between the slope and the ground is 40° and the slope is 10 m long.
a) What is its PE at the top?
b) What is its KE at the bottom?
c) What is the work done by friction?
d) What must be the coefficient of friction?
I am having trouble finding the friction. Once i find the friction, i know how to find the coefficient force of friciton using Ffr=mu*Fn.
The Attempt at a Solution
For part a first I found what the height was by doing 10(sin(40))=Height and got 6.43 m. Than i plugged that into PE=mgh... 5(9.8)(6.43) ... getting 315.07J ...
For part b I used KE=1/2(m)(v2) ... 1/2(5)(62) ... getting 90J ...