## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

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.