(a) If the speed of the block is 5.00 m / s when it reaches the bottom of the ramp, what was the work done on it by the friction force?

Question

A 5.00 kg block is released from rest on a ramp that is inclined at an angle of

60.0

°

below the horizontal. The initial position of the block is a vertical distance of 2.00 m above the bottom of the ramp. (a) If the speed of the block is

5.00

 

m

/

s

when it reaches the bottom of the ramp, what was the work done on it by the friction force?

Answer 1

The potential energy at the top of the ramp, in a perfect world, is all converted to kinetic energy when the block gets to the bottom of the ramp. So EP = EK. But, when there is friction, some of the potential energy gets lost (or converted into work) so we can say:

E_P = E_K + W_F

W_F = E_P - E_K

W_F = mgh - 1/2mv²

W_F = (5)(9.8)(2) - 1/2(5)(5²)

W_F = 98 - 62.5

W_F = 35.5 joules