## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 10 - Problems - Page 402: 84

#### Answer

(a) The period of the oscillations is $3.4~s$ (b) Since the person would still have $9790~J$ of kinetic energy at the bottom of the jump, the person would still be traveling downward at a fast speed. Therefore, the 80.0-kg person should not use the same bungee cord.

#### Work Step by Step

(a) The elastic potential energy in the bungee cord at the bottom is equal to the initial gravitational potential energy at the top of the jump. Note that the bungee cord stretches by $17.0~m$. We can find the spring constant $k$: $U_s = U_g$ $\frac{1}{2}kd^2 = mgh$ $k = \frac{2mgh}{d^2}$ $k = \frac{(2)(60.0~kg)(9.80~m/s^2)(50.0~m)}{(17.0~m)^2}$ $k = 203.5~N/m$ We can find the period of the oscillations: $T = 2\pi~\sqrt{\frac{m}{k}}$ $T = 2\pi~\sqrt{\frac{60.0~kg}{203.5~N/m}}$ $T = 3.4~s$ The period of the oscillations is $3.4~s$ (b) The sum of the kinetic energy and the elastic potential energy in the bungee cord at the bottom is equal to the initial gravitational potential energy at the top of the jump. We can find the kinetic energy at the bottom: $K+U_s = U_g$ $K = U_g-U_s$ $K = mgh-\frac{1}{2}kd^2$ $K = (80.0~kg)(9.80~m/s^2)(50.0~m)-\frac{1}{2}(203.5~N/m)(17.0~m)^2$ $K = 9790~J$ Since the person would still have $9790~J$ of kinetic energy at the bottom of the jump, the person would still be traveling downward at a fast speed. Therefore, the 80.0-kg person should not use the same bungee cord.

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