University Physics with Modern Physics (14th Edition)

Published by Pearson

Chapter 7 - Potential Energy and Energy Conservation - Problems - Exercises - Page 230: 7.30

Answer

The speed of the toolbox just as it reaches the edge will be 5.23 m/s.

Work Step by Step

$K_2 + U_2 = K_1 + U_1 + W_f$ $\frac{1}{2}mv^2+0 = 0 + mgh-F_f~d$ $v^2 = \frac{2mgd~sin(\theta)-2F_f~d}{m}$ $v = \sqrt{\frac{2mgd~sin(\theta)-2F_f~d}{m}}$ $v = \sqrt{\frac{(2)(85.0~N)(4.25~m)~sin(36^{\circ})-(2)(22.0~N)(4.25~m)}{85.0~N~/~9.80~m/s^2}}$ $v = 5.23~m/s$ The speed of the toolbox just as it reaches the edge will be 5.23 m/s.

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