A 25 kg bear slides from rest 12 m down a lodge-pole pine tree moving with a spe

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, 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? What is the bear's kinetic energy at the bottom? 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. Wfnction = Kno friction - Kwith friction)

Explanation / Answer

part A :

Gravitational Potential energy at the top U = mgh

U = 25 *9.81 * 12

U = 2943 Joules

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part B : kinetic energy at the bottom KE = 0.5 Mv ^2

KE = 0.5 * 25* 5.6 * 5.6

KE = 392 J

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Part C:

Work DOne by friction = GPE - KE

Wfric = 2943 - 392

Wfric = 2551 J

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