## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 4 - Problems - Page 148: 15

#### Answer

The speed of the block is $0.50~v_f$ at a distance of $0.50~m$ from the top of the incline.

#### Work Step by Step

We can find an expression for the acceleration: $v_f^2 = v_0^2+2ad$ $a = \frac{v_f^2 - v_0^2}{2d}$ $a = \frac{v_f^2 - 0}{(2)(2.0)}$ $a = \frac{v_f^2}{4.0}$ We can find the distance when the velocity is $0.50~v_f$: $(0.50~v_f)^2 = v_0^2+2ad$ $d = \frac{(0.50~v_f)^2-v_0^2}{2a}$ $d = \frac{0.25~v_f^2}{(2)(\frac{v_f^2}{4.0})}$ $d = 0.50~m$ The speed of the block is $0.50~v_f$ at a distance of $0.50~m$ from the top of the incline.

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