Question

A truck with mass $m$ has a brake failure while going down an icy mountain road of constant downward slope angle $\alpha$ (figure). Initially the truck is moving downhill at speed $v_0$. After careening downhill a distance $L$ with negligible friction, the truck driver steers the runaway vehicle onto a runway truck ramp of constant upward slope angle $\beta$. . The truck ramp 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 ramp before coming to a halt?

Diagram

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

Answer 1

The correct option is D $\frac{\left(v_0^2/2g\right)-L\sin \alpha }{\sin \beta +{\mu }_r\cos \beta }$

Let the distance the truck moves up the ramp by $x$.
Kinetic energy of the truck on icy road is given by,
$K_1=\frac{1}{2}m{v}_0^2$
Potential Energy of the truck on icy road is given by,
$U_1=mgL\sin \alpha$
Kinetic energy of the truck on tuck ramp is given by,
$K_2=0$
Potential Energy of the truck-on-truck ramp is given by,
$U_2=mgx\sin \beta$
Work done is given by,
$W_{other}=-{\mu }_{r}mgx\cos \theta$
Hence, by using work- energy theorem,
$W_{other}=(K_2+U_2)(K_1+U_1)$
Therefore, by putting the values we get,
$(\frac{1}{2}m{v}_0^2+mgL\sin \alpha )(0+mgx\sin \beta )$
$=-{\mu }_{r}mgx\cos \beta$
$x=\frac{K_1+mgL\sin \alpha }{mg(\sin \beta +{\mu }_r\cos \beta )}$
$x=\frac{\left(v_0^2/2g\right)+L\sin \alpha }{\sin \beta +{\mu }_r\cos \beta }$

Final Answer: (b)