Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 6 - Work and Energy - Problems - Page 167: 40


The speed of both boxes as they reach B is $7.67m/s$ The ratio of KE of the heavier box to that of the lighter box is 4 to 1.

Work Step by Step

(a) As the box starts going down and reaches B, there is an increase in KE and a decrease in PE. From the principle of energy conservation, we have $$\Delta KE = -\Delta PE$$ $$\frac{1}{2}m\Delta v^2=-mg\Delta h$$ $$\frac{1}{2}\Delta v^2=-g\Delta h (1)$$ We have $g=9.8m/s^2$, $\Delta h=h_B-h_A=-3m$ and $\Delta v=v_{B}^2-v_{A}^2=v_{B}^2$ $$\frac{1}{2}v_{B}^2=29.4$$ $$v_{B}=7.67m/s$$ b) As we see in equation (1), the change in velocity does not depend on the mass of the box. The principle of energy conservation also only applies for conservative forces, meaning the degree of steepness does not influence $\Delta KE$ or $\Delta PE$ or the final $v$. Therefore, the speed the heavier box reaches B is also $7.67m/s$ c) The lighter box has mass $m_l=11kg$ and the heavier box has mass $m_h=44kg$. We see tht $m_h=4m_l$ The formula for kinetic energy is $KE=\frac{1}{2}mv^2$. Comparing $KE$ of the heavier box and lighter box, we have $$\frac{KE_h}{KE_l}=\frac{\frac{1}{2}m_hv^2}{\frac{1}{2}m_lv^2}=\frac{m_h}{m_l}=4$$
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