Question

A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

Answer 1

Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If  the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N