A truck with mass m has a brake failure while going down an icy mountain road of constant downward slope angle $α$ (see figure). Initially, the truck is moving downhill at speed $v_{0}$ . After travelling downhill a distance L with negligible friction, the truck driver steers the runaway vehicle onto a runway truck ramp of constant upward slope angle $β$ .The truck ramp has a soft sand surface for which the coefficient of rolling friction is 14. What is the distance that the truck moves up the ramp before coming to a halt?

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$\frac{(\frac{{v^{2}}_{0}}{2 g}) - Lsinα}{\sin β + μ_{r} cosβ}$

$\frac{(\frac{{v^{2}}_{0}}{2 g}) + Lsinα}{\sin β + μ_{r} cosβ}$

$\frac{(\frac{{v^{2}}_{0}}{2 g}) + Lsin α}{\sin β - μ_{r} \cos β}$

None of these

answer is A.

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Detailed Solution

Denote the distance the truck moves up the ramp by x. $K_{1} = \frac{1}{2} {mv}^{2}_{0}, U_{1} = mgL \sin α, K_{2} = 0, U_{2} = mgx \sin β and W_{other} = - μ_{r} mgx cosβ . From$

$W_{other} = (K_{2} + U_{2}) - (K_{1} + U_{1}), and solving for x,$

$x = \frac{K_{1)} + mgLsinα}{mg (\sin β + μ_{r} cosβ)} = \frac{(\frac{{v^{2}}_{0}}{2 g}) + Lsin α}{\sin β + μ_{r} \cos β}$

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