## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 6 - Problems - Page 230: 58

#### Answer

(a) The effective spring constant is $6.0\times 10^{10}~N/m$ (b) The block is compressed $8.0~nm$ (c) The work done by the applied forces is $1.92\times 10^{-6}~J$

#### Work Step by Step

(a) We can find the effective spring constant: $F = kx$ $k = \frac{F}{x}$ $k = \frac{120~N}{2.0\times 10^{-9}~m}$ $k = 6.0\times 10^{10}~N/m$ The effective spring constant is $6.0\times 10^{10}~N/m$ (b) We can find the distance that the block is compressed: $x = \frac{F}{k} = \frac{480~N}{6.0\times 10^{10}~N/m} = 8.0\times 10^{-9}~m = 8.0~nm$ The block is compressed $8.0~nm$ (c) We can find the elastic potential energy stored in the block when it is compressed: $U_s = \frac{1}{2}kx^2$ $U_s = \frac{1}{2}(6.0\times 10^{10}~N/m)(8.0\times 10^{-9}~m)^2$ $U_s = 1.92\times 10^{-6}~J$ The work done by the applied forces is $1.92\times 10^{-6}~J$

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