## Engineering Mechanics: Statics & Dynamics (14th Edition)

$t=46.2min$
We can determine the required time as follows: $v=35mi/h(\frac{5280ft/min}{3600s/h})=51.33ft/s$ The kinetic energy is given as $KE=\frac{1}{2}mv^2=\frac{1}{2}(\frac{5000}{32.2})(51.33)^2=204.59\times 10^3lb\cdot ft$ We know that $P=(100W)(\frac{(550lb\cdot ft/s)/hp}{746W/hp})=73.37lb\cdot ft/s$ Now $t=\frac{KE}{P}$ We plug in the known values to obtain: $t=\frac{204.59\times 10^3}{73.73}=2774.98s=46.2min$