Question

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 m s', because of friction. How
much work is done by friction?
(a. 31.3 ms',b. -8840 J)​

Answer 1

$\huge\tt\blue{♬•Answer:}$

(a)

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

(b)

Energy at the top = 9840 J

Energy at the bottom = 1/2 mv² = 1000 J

Now, 9840 + work done by friction = 1000

WORK DONE by friction=1000-9840=-8840 j of work...