Question

A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 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

Thus the force of friction is 235 N

Step-by-step explanation:

When the bear was at the height of 14 m . Its potential energy = m g h

here m is the mass of bear , g is acceleration due to gravity and h is the height .

Thus P.E = 27 x 10 x 14 = 3780 J

The K.E of the bear just before hitting =
`(1)/(2)` m v²

=
`(1)/(2)` x 27 x ( 6.1 )² = 490 J

The force of friction f = P.E - K.E = 3290 J

Because the work done = Force x Distance

Thus frictional force =
`(3290)/(14)` = 235 N