College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 9 - Problems - Page 366: 93


The spring constant is $12.5~N/m$

Work Step by Step

We can find the mass $M$ of the block: $M = \rho~V$ $M = (800~kg/m^3)(0.040~m)^3$ $M = 0.0512~kg$ We can find the weight of the block: $weight = Mg = (0.0512~kg)(9.80~m/s^2) = 0.5018~N$ The weight of the block is $0.5018~N$ According to Archimedes' principle, the buoyant force is equal to the weight of the displaced water. We can find the buoyant force: $F_b = m_w~g$ $F_b = \rho_w~V~g$ $F_b = (1000~kg/m^3)(0.040~m)^3(9.80~m/s^2)$ $F_b = 0.6272~N$ We can find the spring force $F_s$ when the block is submerged in water: $\sum F = 0$ $F_s+Mg-F_b = 0$ $F_s = F_b-Mg$ $F_s = (0.6272~N)-(0.5018~N)$ $F_s = 0.1254~N$ We can find the spring constant: $kx = F_s$ $k = \frac{F_s}{x}$ $k = \frac{0.1254~N}{0.010~m}$ $k = 12.5~N/m$ The spring constant is $12.5~N/m$
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