## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 10 - Energy - Exercises and Problems - Page 274: 42

#### Answer

(a) $h = \frac{k~(\Delta x)^2}{2mg}$ (b) The ice cube goes up to a height of 0.26 m

#### Work Step by Step

(a) The block's potential energy at the maximum height will be equal to the initial energy stored in the spring. We can find an expression for the maximum height $h$; $PE = U_s$ $mgh = \frac{1}{2}k(\Delta x)^2$ $h = \frac{k~(\Delta x)^2}{2mg}$ (b) We can use the expression from part (a) to find the maximum height $h$; $h = \frac{k~(\Delta x)^2}{2mg}$ $h = \frac{(25~N/m)(0.10~m)^2}{(2)(0.050~kg)(9.80~m/s^2)}$ $h = 0.26~m$ The ice cube goes up to a height of 0.26 m.

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