College Physics (4th Edition)

(a) $k = 25.2~N/m$ (b) The jumper's speed is $25.0~m/s$
(a) Let $h_1 = 182~m$ and let $h_2 = 68~m$. Note that the bungee cord is stretched a distance of $84~m$ when the jumper is 68 meters above the bottom. We can use conservation of energy to find the spring constant of the bungee cord: $\frac{1}{2}kx^2+mgh_2 = mgh_1$ $k = \frac{2mg~(h_1-h_2)}{x^2}$ $k = \frac{(2)(780~N)~(182~m-68~m)}{(84~m)^2}$ $k = 25.2~N/m$ (b) Let $h_1 = 182~m$ and let $h_3 = 92~m$. Note that the bungee cord is stretched a distance of $60~m$ when the jumper is 92 meters above the bottom. We can use conservation of energy to find the jumper's speed: $\frac{1}{2}mv^2+\frac{1}{2}kx^2+mgh_3 = mgh_1$ $\frac{1}{2}mv^2 = mg~(h_1-h_3) - \frac{1}{2}kx^2$ $v^2 = \frac{2mg~(h_1-h_3) - kx^2}{m}$ $v = \sqrt{\frac{2mg~(h_1-h_3) - kx^2}{m}}$ $v = \sqrt{\frac{(2)(780~N)~(182~m-92~m) - (25.2~N/m)(60~m)^2}{(780~N~/~9.80~m/s^2)}}$ $v = 25.0~m/s$ The jumper's speed is $25.0~m/s$.