Question

Assignment 4.3
FRICTION
Consider a person on a sled sliding down a 100 m long hill on a 30° incline. The
mass is 20 kg, and the person has a velocity of 2 ms'down the hill when they're
at the top. (a) How fast is the person traveling at the bottom of the hill?
(b) If, the velocity at the bottom of the hill is 10 ms', because of friction. How
much work is done by friction?​

Answer 1

Explanation:

At the top: PE = mgh = (20) (9.8) (100sin30°) = 9800 J

KE = 1/2 mv2 = 1/2 (20) (2)2 = 40 J

Total mechanical energy at the top = 9800 + 40 = 9840 J

At the bottom: PE = 0 KE = 1/2 mv2

Total mechanical energy at the bottom = 1/2 mv2

If we conserve mechanical energy, then the mechanical energy at the top must equal what we have at the bottom. This gives:

1/2 mv2 = 9840, so v = 31.3 m/s.