## Physics: Principles with Applications (7th Edition)

(a) $W_g = 1.0 \times 10^4~J$ (b) $v = 16~m/s$
(a) Note that the angle between the force of gravity (which points straight down) and the cyclist's velocity vector is $86.0^{\circ}$. We can find the work done by gravity on the cyclist. $W_g = F\cdot d~cos(86.0^{\circ})$ $W_g = mg \cdot d~cos(86.0^{\circ})$ $W_g = (85~kg)(9.80~m/s^2)(180~m)~cos(86.0^{\circ})$ $W_g = 10,000~J$ (b) The cyclist's kinetic energy after coasting down the hill will be equal to the work done on the cyclist by gravity. $KE = W_g$ $\frac{1}{2}mv^2 = W_g$ $v^2 = \frac{2~W_g}{m}$ $v = \sqrt{\frac{2~W_g}{m}}$ $v = \sqrt{\frac{(2)(10,000~J)}{85~kg}}$ $v = 16~m/s$