College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 6 - Problems - Page 233: 100

Answer

(a) The pitcher does $124~J$ of work on the baseball. (b) A pitcher would have to throw 10,300 fastballs in order to burn off a 1520 kilocalorie meal.

Work Step by Step

(a) We can find the kinetic energy of the ball: $\frac{1}{2}mv^2 = \frac{1}{2}(0.153~kg)(40.2~m/s)^2 = 124~J$ The pitcher does $124~J$ of work on the baseball. (b) This energy is only 20% of the chemical energy that is used to throw the baseball. We can find the amount of chemical energy $E_c$ that is used: $0.20~E_c = 124~J$ $E_c = \frac{124~J}{0.20}$ $E_c = 620~J$ We can convert this energy to units of kilocalories: $E_c = 620~J \times \frac{1~kcal}{4184~J} = 0.148~kcal$ We can find the number of fastballs a pitcher would have to throw in order to burn off a 1520 kilocalorie meal: $\frac{1520~kcal}{0.148~kcal/fastball} = 10,300~fastballs$ A pitcher would have to throw 10,300 fastballs in order to burn off a 1520 kilocalorie meal.
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