University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 7 - Potential Energy and Energy Conservation - Problems - Exercises - Page 233: 7.65

Answer

(a) The speed when the block loses contact with the spring is 0.747 m/s. (b) The maximum speed of the block is 0.930 m/s.

Work Step by Step

(a) $K = U_s + W_f$ $\frac{1}{2}mv^2 = \frac{1}{2}kx^2-mg~\mu_k~x$ $v^2 = \frac{kx^2-2mg~\mu_k~x}{m}$ $v = \sqrt{\frac{kx^2-2mg~\mu_k~x}{m}}$ $v = \sqrt{\frac{(45.0~N/m)(0.280~m)^2-(2)(1.60~kg)(9.80~m/s^2)(0.300)(0.280~m)}{1.60~kg}}$ $v = 0.747~m/s$ The speed when the block loses contact with the spring is 0.747 m/s. (b) The maximum speed will occur at the point where the force from the spring is equal in magnitude to the force of friction. $kx = mg~\mu_k$ $x = \frac{mg~\mu_k}{k}$ $x = \frac{(1.60~kg)(9.80~m/s^2)(0.300)}{45.0~N/m}$ $x = 0.1045~m$ We can use this value of $x$ to find the maximum speed of the block. $\frac{1}{2}mv^2 = \frac{1}{2}mv_{max}^2 +\frac{1}{2}kx^2 - mg~\mu_k~x$ $\frac{1}{2}mv_{max}^2 = \frac{1}{2}mv^2 -\frac{1}{2}kx^2 + mg~\mu_k~x$ $v_{max}^2 = \frac{mv^2 -kx^2 + 2mg~\mu_k~x}{m}$ $v_{max} = \sqrt{\frac{(1.60~kg)(0.747~m/s)^2 -(45.0~N/m)(0.1045~m)^2 + (2)(1.60~kg)(9.80~m/s^2)(0.300)(0.1045~m)}{1.60~kg}}$ $v_{max} = 0.930~m/s$
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