Answer
2.6 $\times 10^{2}$ m
Work Step by Step
Since the ramp is frictionless, the mechanical energy $E_{mec}$ is conserved and so $ΔE_{mec} = ΔK + ΔU= \frac{1}{2} m(v^2–v_{0}^{2})+mgΔh=0$.
Making a right triangle with the ramp with length L, the change in height the truck would undergo when going up the ramp would be $\Delta h = L sin \theta$. Plugging in this expression and solving for L, we get:
$ \frac{1}{2} m(v^2–v_{0}^{2})+mgΔh=0$.
$ \frac{1}{2} m(v^2–v_{0}^{2})+mgL sin \theta=0$.
$mgL sin \theta = – \frac{1}{2} m(v^2–v_{0}^{2})$
$L = – \frac{ v^2–v_{0}^{2}}{2gsin \theta}$
Now substituting $\theta$ = $15 ^{\circ}$,$ v_{0}$ = 130 km/h = 36.1 m/s, v = 0 and g = 9.8 m/s²:
$L = – \frac{ 0^2–(36.1 m/s)^{2}}{2(9.8 m/s²) sin 15^{\circ}}$
= 257.06 m $\approx$ = 2.6 $\times 10^{2}$
