Question

A track has two inclined surface $AB$ and $DC$ each of length $3\, m$ and angle of inclination of $30^°$ with the horizontal and a central horizontal part of length $4\, m$ as shown in figure. A block of mass $0.2\, kg$ 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 ?

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

Answer 1

Answer: (A)

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 = 3 \,m \times sin\, 30^° = 1.5\, m$. 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 distance travelled on the horizontal part, the work done against friction is

From the principle of conservation of energy, we have
$mgh = \mu mgs$
$s = \frac{h}{\mu} = \frac{1.5\,m}{0.2} = 7.5\,m$
$ = 750 \times 10^{-2}\,m$
i.e.y the block travels a total distance of $7.5\, m$ on the horizontal track $BC$ before coming to rest, starting from $B$. It travels $4 \,m$ from $B$ to $C$ and $3.5\, m$ from $C$ to a point $P$ where it comes to rest. Hence the block comes to rest at point $P$ at a distance of $0.5\, m$ from $B$.