## University Physics with Modern Physics (14th Edition)

Published by Pearson

# Chapter 7 - Potential Energy and Energy Conservation - Problems - Exercises - Page 230: 7.29

#### Answer

(a) The speed at the bottom of the hill is 8.16 m/s. (b) The internal energy generated when crossing the rough patch was 766 J.

#### Work Step by Step

$K_2 + U_2 = K_1 + U_1 + W_f$ $\frac{1}{2}mv_2^2+0 = \frac{1}{2}mv_1^2+ mgh - mg~\mu_k~d$ $v_2^2 = v_1^2+ 2gh - 2g~\mu_k~d$ $v_2 = \sqrt{v_1^2+ 2gh - 2g~\mu_k~d}$ $v_2 = \sqrt{(6.50~m/s)^2+ (2)(9.80~m/s^2)(2.50~m) - (2)(9.80~m/s^2)(0.300)(4.20~m)}$ $v_2 = 8.16~m/s$ The speed at the bottom of the hill is 8.16 m/s. (b) The internal energy is equal in magnitude to the work done by friction when crossing the rough patch. $W_f = -mg~\mu_k~d$ $W_f = -(62.0~kg)(9.80~m/s^2)(0.300)(4.20~m)$ $W_f = -766~J$ The internal energy generated when crossing the rough patch was 766 J.

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