## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 6 - Problems - Page 229: 40

#### Answer

Both crates are moving with a speed of $2.91~m/s$ after the crates have moved a distance of 1.4 meters.

#### Work Step by Step

The kinetic energy of the system of two crates will be equal in magnitude to the net change in the potential energy of the two crates. Note that crate $m_2$ falls a distance of 1.4 meters as crate $m_1$ moves up the incline a distance of 1.4 meters: $\frac{1}{2}(m_1+m_2)~v^2 = m_2gh_2-m_1gh_1$ $v^2 = \frac{2g~(m_2h_2-m_1h_1)}{m_1+m_2}$ $v = \sqrt{\frac{2g~(m_2h_2-m_1h_1)}{m_1+m_2}}$ $v = \sqrt{\frac{(2)(9.80~m/s^2)~[(16.3~kg)(1.4~m)-(12.4~kg)(1.4~m)~sin~36.9^{\circ}]}{16.3~kg+12.4~kg}}$ $v = 2.91~m/s$ Both crates are moving with a speed of $2.91~m/s$ after the crates have moved a distance of 1.4 meters.

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