91Ó°ÊÓ

Let be the mass of the truck.

We take $h=0$ at the bottom of the icy road and the beginning of the truck ramp.

Then we have,

Case 1: The truck moves on the icy road of a slope $\mathrm{Î\pm }$ and distance $L$before the beginning of the truck ramp.

Let us take point 1 at the initial point and point 2 at the bottom of the road.

Then we have,

$\begin{array}{l}{h}_1=L\sin \mathrm{Î\pm },\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}{v}_1=v_0\\ h_2=0,\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰}{v}_2=v\text{ â¶Ä‰}\end{array}$

Case 2: The truck moves on the truck ramp of slope angle $\mathrm{β}$ and distance$d$ until it stops.

Let us take point 1 at the beginning of the ramp and point 2 at the halt point.

Then we have,

$\begin{array}{l}{h}_1=0,\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰}{v}_1=v\\ h_2=d\sin \mathrm{β},\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰}{v}_2=v\text{ â¶Ä‰}\end{array}$

Since there are other forces acting other than gravity, work-energy theorem is given by,

$K_i+U_i+W=K_f+U_f\text{ â¶Ä‰â€‰}⋯⋯\left(1\right)$,

where the kinetic energy is given by,

$K=\frac{1}{2}m{v}^2\text{ â¶Ä‰}⋯⋯\left(2\right)$

And the gravitational potential energy is given by,

.$U=mgh\text{ â¶Ä‰â€‰}⋯⋯\left(3\right)$

We know that the friction force and resistance are always acting opposite to the direction of motion.

Thus, the work done is,

,$W_f=−fâ‹\dots s\text{ â¶Ä‰â€‰}⋯⋯\left(4\right)$

Where, $s$ is the displacement.

The kinetic friction force is,

,$f_k={\mathrm{μ}}_kâ‹\dots n\text{ â¶Ä‰â€‰}⋯⋯\left(5\right)$

Where $n$ is the normal force.

Let us calculate energy quantities for case 1.

Putting $v_1$in $\left(2\right)$, we get,

.$K_1=\frac{1}{2}m{v}_0^2$

Also, substituting $\text{ }{h}_1$in $\left(3\right)$, we get,

.$U_1=mgL\sin \mathrm{Î\pm }$

Now, we substitute $v_2$in $\left(2\right)$, we get,

.$K_2=\frac{1}{2}m{v}^2\text{ â¶Ä‰}$

Since truck ends at 0 potential level, we get,

.$U_2=0$

Since the icy road is frictionless, the work done is,

.$W=0$

Substituting all these values of energy and work quantities in , we get,

$\begin{array}{l}\frac{1}{2}m{v}_0^2+mgL\sin \mathrm{Î\pm }+0=\frac{1}{2}m{v}^2+0\\ ⇒v^2=v_0^2+2gL\sin \mathrm{Î\pm }\text{ â¶Ä‰â€‰}⋯⋯\left(6\right)\end{array}$

Let us calculate energy quantities for case 2.

Putting $v_1$in $\left(2\right)$, we get,

.$K_1=\frac{1}{2}m{v}^2$

Also, substituting $\text{ }{h}_2$in $\left(3\right)$, we get,

.$U_2=mgd\sin \mathrm{β}$

Since the truck comes to rest, we get,

$K_2=0$.

Since truck startsfrom the 0 potential level, we get,

.$U_1=0$

Since the icy road is frictionless, the work done is,

.$W=0$

Applying Newton’s Second Law, to the truck along the direction perpendicular to the ramp, we get,

$\begin{array}{l}∑F=n−mg\cos \mathrm{β}=0\\ →n=mg\cos \mathrm{β}\end{array}$

Substituting this in $\left(5\right)$we get,

$\begin{array}{l}{f}_k={\mathrm{μ}}_râ‹\dots \left(mg\cos \mathrm{β}\right)\\ ⇒f_k={\mathrm{μ}}_{r}mg\cos \mathrm{β}\end{array}$

Now, substituting in $\left(4\right)$we get,

.$W=−mgd{\mathrm{μ}}_r\cos \mathrm{β}$

Substituting all these values of energy and work quantities in $\left(1\right)$, we get,

$\begin{array}{l}\frac{1}{2}m{v}^2+0−mgd\cos \mathrm{β}=0+mgd\sin \mathrm{β}\\ ⇒v^2=2gd(\sin \mathrm{β}+{\mathrm{μ}}_r\cos \mathrm{β})\text{ â¶Ä‰}\end{array}$

Using the value from $\left(6\right)$ we get,

$\begin{array}{l}{v}_0^2+2gL\sin \mathrm{Î\pm }\text{ â¶Ä‰}=2gd(\sin \mathrm{β}+{\mathrm{μ}}_r\cos \mathrm{β})\text{ â¶Ä‰}\\ ⇒d=\frac{v_0^2+2gL\sin \mathrm{Î\pm }}{2g(\sin \mathrm{β}+{\mathrm{μ}}_r\cos \mathrm{β})}\end{array}$

Hence, the distance that the truck moves up the ramp is,$d=\frac{v_0^2+2gL\sin \mathrm{Î\pm }}{2g(\sin \mathrm{β}+{\mathrm{μ}}_r\cos \mathrm{β})}$ .