## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 6 - Problems - Page 233: 97

#### Answer

The object slides a distance of 0.33 meters up the incline.

#### Work Step by Step

The gravitational potential energy at the highest point will be equal to the elastic potential energy stored in the spring initially. Note that the maximum height $h$ is equal to $d~sin~\theta$, where $d$ is the distance the object slides up the incline. We can find the distance $d$: $mgh = \frac{1}{2}kx^2$ $mgd~sin~\theta = \frac{1}{2}kx^2$ $d = \frac{kx^2}{2mg~sin~\theta}$ $d = \frac{(40.0~N/m)(0.20~m)^2}{(2)(0.50~kg)(9.80~m/s^2)~sin~30.0^{\circ}}$ $d = 0.33~m$ The object slides a distance of 0.33 meters up the incline.

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