#### Answer

The length of the ramp should be 120 meters.

#### Work Step by Step

We can find the initial kinetic energy of the truck.
$KE = \frac{1}{2}mv^2$
$KE = \frac{1}{2}(15,000~kg)(35~m/s)^2$
$KE = 9.19\times 10^6~J$
Let $d$ be the distance the truck travels along the ramp. We can find an expression for the height $h$.
$\frac{h}{d} = sin(\theta)$
$h = d~sin(\theta)$
The final potential energy will be equal to the sum of the initial kinetic energy and the work done by friction.
$PE = KE+W_f$
$PE - W_f = KE$
$mgh - (-mg~\mu_k~d) = KE$
$mgd~sin(\theta) +mg~\mu_k~d = KE$
$d = \frac{9.19\times 10^6~J}{(15,000~kg)(9.80~m/s^2)~[sin(6.0^{\circ}) +0.40]}$
$d = 120~m$
The length of the ramp should be 120 meters.