## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 6 - Problems - Page 231: 69

#### Answer

(a) $v = 2.21~m/s$ (b) The spring is compressed a distance of $20.9~cm$ (c) The block rises up to its original height of $0.50~m$

#### Work Step by Step

(a) We can use conservation of energy to find the speed when the block is at a height of $0.25~m$: $\frac{1}{2}mv^2+mgh_2 = mgh_1$ $v^2 = 2g~(h_1-h_2)$ $v = \sqrt{2g~(h_1-h_2)}$ $v = \sqrt{(2)(9.80~m/s^2)~(0.50~m-0.25~m)}$ $v = 2.21~m/s$ (b) By conservation of energy, the elastic potential energy stored in the spring will be equal to the initial gravitational potential energy: $\frac{1}{2}kx^2 = mgh$ $x^2 = \frac{2mgh}{k}$ $x = \sqrt{\frac{2mgh}{k}}$ $x = \sqrt{\frac{(2)(2.0~kg)(9.80~m/s^2)(0.50~m)}{450~N/m}}$ $x = 0.209~m = 20.9~cm$ The spring is compressed a distance of $20.9~cm$ (c) When the block goes back to the left, by conservation of energy, the block rises up to its original height of $0.50~m$

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