A truck with mass `m` has a brake failure while going down an icy mountain road of constant downwards slope angle `alpha` (see figure) . Initial , the truck is moving downhill at speed `v_(0)` . After carrying downhill a distance `L` with negligible friction for the truck driver steers the runaway vehicle onto a runway truck rump of constant upward slope angle `Beta` . The truck rump has to soft sand surface for which the coefficient of rolling friction is `mu_(r)`. What is the distance that the truck moves up the rump before coming to a halt?

Options

A. `((v_(0)^(2)//2g) + L sin alpha)/(sin beta + mu_(r), cos beta)`
B. `((v_(0)^(2)g) - L sin alpha)/((sin beta + mu_(r), cos beta)`
C. `((v_(0)^(2)//2g) + L sin alpha)/(sin beta - mu_(r), cos beta)`
D. None of these

Answer

A

Explanation

Denote the distance the truck moves up the ramp by
`x. K_(1)=(1)/(2)mv_(0)^(2), U_(1) =mgL sin alpha, K_(2) =0, U_(2) =mgx`
sin beta` and `W_(other =- mu_(r) mgx cos beta`. From
`x=(K_(1)+mgL sin beta)/(mg (sin beta +mu_(r) cos beta))=((v_(0)^(2)//2g)+L sin beta)/(sin beta + mu_(r) cos beta)`.