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 211: 104b

Answer

The speed when it passes through the origin is $~~6.4~m/s$

Work Step by Step

We can find the potential energy at $~~x = 5.0~m$: $\Delta U = -\int_{0}^{5.0}~F(x)~dx$ $\Delta U = -\int_{0}^{5.0}~(-3.0~x-5.0~x^2)~dx$ $\Delta U = \int_{0}^{5.0}~(3.0~x+5.0~x^2)~dx$ $\Delta U = (1.5~x^2+\frac{5.0~x^3}{3})\Big \vert_{0}^{5.0}$ $\Delta U = [1.5~(5.0)^2+\frac{(5.0)~(5.0)^3}{3}]-0$ $\Delta U = 245.8~J$ Since $~~U=0~~$ at $~~x = 0,~~$ then $~~U = 245.8~J~~$ at $~~x = 5.0~m$ We can use conservation of energy to find the speed when it passes through the origin: $K_f+U_f = K_0+U_0$ $K_f+0 = K_0+U_0$ $\frac{1}{2}mv_f^2 = \frac{1}{2}mv_0^2+245.8~J$ $v_f^2 = v_0^2+\frac{(2)(245.8~J)}{m}$ $v_f^2 = v_0^2+\frac{(2)(245.8~J)}{20~kg}$ $v_f = \sqrt{v_0^2+\frac{(2)(245.8~J)}{20~kg}}$ $v_f = \sqrt{(-4.0~m/s)^2+\frac{(2)(245.8~J)}{20~kg}}$ $v_f = 6.4~m/s$ The speed when it passes through the origin is $~~6.4~m/s$
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