## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 5 - Problems - Page 187: 41

#### Answer

(a) The tension in the string at the bottom of the swing is $13~N$ (b) As well as balancing out the downward gravitational force on the bob, the tension must also provide the required centripetal force for the bob.

#### Work Step by Step

(a) We can find the tension at the bottom of the swing: $\sum F = \frac{mv^2}{r}$ $T-mg = \frac{mv^2}{r}$ $T = mg + \frac{mv^2}{r}$ $T = (1.0~kg)(9.80~m/s^2) + \frac{(1.0~kg)(1.6~m/s)^2}{0.80~m}$ $T = 13~N$ The tension in the string at the bottom of the swing is $13~N$ (b) The tension is greater than the weight of the bob because the tension is equal in magnitude to the sum of the weight of the bob and the required centripetal force to keep the bob moving in a circular path. That is, as well as balancing out the downward gravitational force on the bob, the tension must also provide the required centripetal force for the bob.

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