Question

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 v₀. After careening downhill a distance L with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle β. The truck ramp has a soft sand surface for which the coefficient of rolling friction is μ_r. What is the distance that the truck moves up the ramp before coming to a halt?

• A. $({v_0}^2/2g)+L\sin \alpha /\sin \beta +{\mu }_r\cos \beta$
• B. $({v_0}^2/2g)-L\sin \alpha /\sin \beta +{\mu }_r\cos \beta$
• C. $({v_0}^2/2g)+L\sin \alpha /\sin \beta -{\mu }_r\cos \beta$
• D. None of these.

Answer 1

The correct option is A

$({v_0}^2/2g)+L\sin \alpha /\sin \beta +{\mu }_r\cos \beta$

Explanation for the correct option

Step 1. Given data

Mass of the truck $=m$

Initial velocity $=v_0$

Downhill distance $=L$

Coefficient of rolling friction $={\mu }_r$

Downhill slope $=\alpha$

Uphill slope $=\beta$

Step 2. Calculation of distance travelled by the ramp uphill before coming to halt (x)

The various assumptions and forces acting on the truck are shown in the diagram below.

From triangle AOC, $\sin \alpha =h/L$

$h=L\sin \alpha$……….(a)

From triangle BOD, $\sin \beta =h_1/x$

$h_1=x\sin \beta$……….(b)

From the law of conservation of energy, total energy at A and B must be the same.

$mgh+1/2m{v^2}_0=mg{h}_1+{\mu }_{r}Nx$………………..(C)

Substituting the values of h and h₁ in equation (c), we get

$mgL\sin \alpha +1/2m{v_0}^2=mgx\sin \beta +{\mu }_{r}Nx$…………….(d)

From figure (B), $N=mg\cos \beta$

Using the above value in equation (d), we get

$$\begin{gathered} mgL\sin \alpha +1/2m{v_0}^2=mgx\sin \beta +{\mu }_{r}mg\cos \beta x \\ m(gL\sin \alpha +1/2m{v_0}^2)=x(mg\sin \beta +{\mu }_{r}mg\cos \beta ) \\ x=(gl\sin \alpha +1/2m{v_0}^2)/(g\sin \beta +{\mu }_{r}g\cos \beta ) \\ x=({v_0}^2/2g+L\sin \alpha )/(\sin \beta +{\mu }_r\cos \beta ) \end{gathered}$$

Hence, option (a) will be correct.