Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 6 - Work and Energy - Problems - Page 166: 45

Answer

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Work Step by Step

a. The total work done by the cyclist, against gravity, equals the change in potential energy for the entire climb. $$W=\Delta PE=mg(h_f-h_i)$$ $$=(75.0kg)(9.80m/s^2)(125m)=9.19\times10^4 J$$ b. The work done by the average tangential force on the pedals during one revolution is the product of the force and the circumference. This equals the change in PE for one revolution of the pedals. We see that a vertical rise $\Delta h$ is related to the distance D traveled up the incline by $\Delta h=D sin\theta$. $$W_{pedal}=F2\pi r=\Delta PE_{1\;rev}=mg\Delta h=mg (5.10m)sin7.5^{\circ}$$ Solve for the average tangential force. $$F=\frac{ mg (5.10m)sin7.5^{\circ}}{2\pi r }$$ $$F=\frac{(75.0kg)(9.80m/s^2) (5.10m)sin7.5^{\circ}}{2\pi (0.180m)}=433N$$
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