#### Answer

$t=46.2min$

#### Work Step by Step

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:
$t=\frac{204.59\times 10^3}{73.73}=2774.98s=46.2min$