Question

A track has two inclined surface $AB$ and $DC$ each of length $3m$ and angle of inclination of ${30}^{\circ }$ with the horizontal and a central horizontal part of length $4m$ as shown in figure. A block of mass $0.2kg$ slides from rest from point $A$. The inclined surfaces are frictionless. If the coefficient of friction between the block and the horizontal flat surface is $0.2$, where will the block finally come to rest? [in ${10}^{-1}m]$

• A. $0.5m$ away from point $B$
• B. $3.5m$ away from point $B$
• C. $0.5m$ away from point $C$
• D. $1.5m$ away from point $C$

Answer 1

The correct option is A $0.5m$ away from point $B$

At point $A$, the energy of the block is entirely potential $=mgh$,

where $h$ is the height of $A$ above the horizontal surface $=AB\sin \theta =3m\times \sin {30}^{\circ }=1.5m$.

When the block is released, it reaches point $B$ where the entire potential energy is converted into kinetic energy. It will, move on track $BC$. On reaching $C$, it will rise on the surface $CD$ to a point $E$ and will then return to $C$ moving towards $B$ and the process goes on until the block loses all its energy and comes to rest.

The body loses energy while moving on the horizontal part $BC$ of the track. If $s$ is the total distantance travelled on the horizontal part, the work done against friction is
$W=\mu R\times s=\mu mgs$.

From the principle of conservation of energy,

we have $mgh=\mu mgs$
$s=\frac{h}{\mu }=\frac{1.5m}{0.2}=7.5m=750\times {10}^{-2}m$

i.e., the block travels a total distance of $7.5m$ on the horizontal track $BC$ before coming to rest, starting from $B$. It travels $4m$ from $B$ to $C$ and $3.5m$ from $C$ to a point $P$ where it comes to rest. Hence the block comes to rest at point $P$ at a distance of $0.5m$ from $B$.