Question

A 25 kg bear slides from rest 12 m down a lodge-pole pine tree moving with a speed of 5.6 m/s at the bottom. (g = 9.8 m/s2.)

a. What is the gravitational potential energy of the bear at the top, if the bottom of the tree is assumed to have zero gravitational potential energy?

b.What is the bear’s kinetic energy at the bottom?

c.  Determine the work done by friction, by calculating what the bear’s kinetic energy would be with no friction and subtracting the kinetic energy you found in part b (i.e. Wfriction = Kno friction – Kwith friction).

Answer 1

(a)
Gravitational potential energy of the bear at the top = m*g*h
Gravitational potential energy of the bear at the top = 25 * 9.8 * 12 J
Gravitational potential energy of the bear at the top = 2940 J

(b)
Kinetic energy at the bottom = 1/2 * m*v^2
Kinetic energy at the bottom = 1/2 * 25 * 5.6^2
Kinetic energy at the bottom = 392 J

(c)
If there was no friction, Kinetic Energy would have been , = 2940 J
Work done by Friction = 2940 - 392
Work done by Friction = 2548 J