## University Physics with Modern Physics (14th Edition)

Published by Pearson

# Chapter 7 - Potential Energy and Energy Conservation - Problems - Exercises - Page 234: 7.71

#### Answer

At the top of the path, the magnitude of the normal force is 0.456 N.

#### Work Step by Step

Let point A be the bottom of the loop. We can find the speed at point A. $\frac{mv_A^2}{R} = \sum F$ $\frac{mv_A^2}{R} = F_N-mg$ $v_A^2 = \frac{R~(F_N-mg)}{m}$ $v_A = \sqrt{\frac{R~(F_N-mg)}{m}}$ $v_A = \sqrt{\frac{(0.800~m)[3.40~N-(0.0500~kg)(9.80~m/s^2)]}{0.0500~kg}}$ $v_A = 6.82~m/s$ Let point B be the top of the loop. We can find the speed at point B. $K_B+U_B = K_A+U_A$ $K_B = K_A+0 - U_B$ $\frac{1}{2}mv_B^2 = \frac{1}{2}mv_A^2-2mgR$ $v_B^2 = v_A^2-4gR$ $v_B = \sqrt{v_A^2-4gR}$ $v_B = \sqrt{(6.82~m/s)^2-(4)(9.80~m/s^2)(0.800~m)}$ $v_B = 3.89~m/s$ We can find the normal force $F_N$ at point B. $\sum F = \frac{mv_B^2}{R}$ $F_N+mg = \frac{mv_B^2}{R}$ $F_N = \frac{mv_B^2}{R}-mg$ $F_N = \frac{(0.0500~kg)(3.89~m/s)^2}{0.800~m}-(0.0500~kg)(9.80~m/s^2)$ $F_N = 0.456~N$ At the top of the path, the magnitude of the normal force is 0.456 N.

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