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

Published by Pearson

# Chapter 7 - Potential Energy and Energy Conservation - Problems - Exercises - Page 231: 7.37

#### Answer

(a) The force of friction on the box is 637 N. Since the system is at rest, the force of friction on the bag of gravel is zero. (b) The speed of the bucket will be 2.99 m/s

#### Work Step by Step

(a) We can find the maximum possible force of static friction on the box when the bag of gravel is on the box. $F_f = mg~\mu_s = (130.0~kg)(9.80~m/s^2)(0.700)$ $F_f = 892~N$ The weight of the bucket is $(65.0~kg)(9.80~m/s^2)$, which is 637 N. Since the maximum possible force of static friction on the box is greater than the weight of the bucket, the system is at rest. The force of friction on the box is 637 N. Since the system is at rest, the force of friction on the bag of gravel is zero. (b) Let $m_1$ be the mass of the bucket. Let $m_2$ be the mass of the box. $K_2+U_2 = K_1+U_1+W_f$ $\frac{1}{2}(m_1+m_2)v^2 = 0 + m_1gh - m_2g~\mu_k~d$ $v^2 = \frac{2m_1gh - 2m_2g~\mu_k~d}{m_1+m_2}$ $v = \sqrt{\frac{2m_1gh - 2m_2g~\mu_k~d}{m_1+m_2}}$ $v = \sqrt{\frac{(2)(65.0~kg)(9.80~m/s^2)(2.00~m) - (2)(80.0~kg)(9.80~m/s^2)(0.400)(2.00~m)}{65.0~kg+80.0~kg}}$ $v = 2.99~m/s$ The speed of the bucket will be 2.99 m/s, We can use Newton's laws to check this answer. $ma = \sum F$ $a = \frac{m_1g - m_2g~\mu_k}{m_1+m_2}$ $a = \frac{(65.0~kg)(9.80~m/s^2) - (80.0~kg)(9.80~m/s^2)(0.400)}{145.0~kg}$ $a = 2.23~m/s^2$ We can use the acceleration to find the speed of the system. $v^2=v_0^2+2ay = 0+2ad$ $v = \sqrt{2ay} = \sqrt{(2)(2.23~m/s^2)(2.00~m)}$ $v = 2.99~m/s$

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