## College Physics (4th Edition)

Published by McGraw-Hill Education

# Review & Synthesis: Chapters 6-8 - Review Exercises - Page 325: 33

#### Answer

(a) The speed of the bike at the bottom of the hill is $~19.4~m/s$ (b) With a less massive rider, the speed at the bottom would be slower.

#### Work Step by Step

(a) Let $M$ be the mass of the bike. Let $m$ be the mass of each wheel. We can find the total kinetic energy of the bike when it is moving at a speed $v$: $KE = KE_{tr}+KE_{rot}$ $KE = \frac{1}{2}Mv^2+2\times \frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}Mv^2+(mR^2)~(\frac{v}{R})^2$ $KE = \frac{1}{2}Mv^2+mv^2$ $KE = (\frac{M}{2}+m)~v^2$ We can use conservation of energy to find the speed of the bike at the bottom of the hill: $KE_f = U_0$ $(\frac{M}{2}+m)~v^2 = Mgh$ $v^2 = \frac{Mgh}{\frac{M}{2}+m}$ $v = \sqrt{\frac{Mgh}{\frac{M}{2}+m}}$ $v = \sqrt{\frac{(80.0~kg)(9.80~m/s^2)(20.0~m)}{\frac{80.0~kg}{2}+1.5~kg}}$ $v = 19.4~m/s$ The speed of the bike at the bottom of the hill is $~19.4~m/s$ (b) If the rider is less massive, then a smaller fraction of the kinetic energy would be in the form of translational kinetic energy. Therefore, with a less massive rider, the speed at the bottom would be slower.

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