Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 8 - Potential Energy and Conservation of Energy - Problems - Page 212: 124b

Answer

The work required to increase the separation of the particles from $~~x = x_1~~$ to $~~x = x_1+d~~$ is $~~\frac{G~m_1~m_2~d}{(x_1)(x_1+d)}$

Work Step by Step

Since $F(x)$ is attractive, note that $F(x) = -\frac{G~m_1~m_2}{x^2}$ We can find the potential energy function: $U(x) = -\int~F(x)~dx$ $U(x) = -\int~(-\frac{G~m_1~m_2}{x^2})~dx$ $U(x) = \int~\frac{G~m_1~m_2}{x^2}~dx$ $U(x) = -\frac{G~m_1~m_2}{x}$ We can find the potential energy when $x = x_1$: $U_1 = -\frac{G~m_1~m_2}{x_1}$ We can find the potential energy when $x = x_1+d$: $U_2 = -\frac{G~m_1~m_2}{x_1+d}$ We can find $\Delta U$: $\Delta U = U_2-U_1$ $\Delta U = (-\frac{G~m_1~m_2}{x_1+d})-(-\frac{G~m_1~m_2}{x_1})$ $\Delta U = \frac{G~m_1~m_2}{x_1}-\frac{G~m_1~m_2}{x_1+d}$ $\Delta U = \frac{(G~m_1~m_2)(x_1+d)}{(x_1)(x_1+d)}-\frac{(G~m_1~m_2)(x_1)}{(x_1)(x_1+d)}$ $\Delta U = \frac{G~m_1~m_2~d}{(x_1)(x_1+d)}$ The work required to increase the separation of the particles from $~~x = x_1~~$ to $~~x = x_1+d~~$ is $~~\frac{G~m_1~m_2~d}{(x_1)(x_1+d)}$
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