Question

A particle slides along a track with elevated ends and a flat central part, as shown in figure. The flat portion $BC$ has a length $l=3.0\text{m}$. The curved portions of the track are frictionless. For the flat part of the track the coefficient of kinetic friction is ${\mu }_k=0.20$, If the particle is released at point $A$ which is at height $h=1.5\text{m}$ above the flat part of the track, then Where does the particle finally comes to rest w.r.t $B$ ?

• A. $2.5\text{m}$
• B. $1.5\text{m}$
• C. $3.5\text{m}$
• D. $4.5\text{m}$

Answer 1

The correct option is B $1.5\text{m}$

As initial mechanical energy of the particle is, $E_i=mgh$.

Final mechanical energy, $E_f=0$

So, loss in mechanical energy$\Delta E=mgh-0=mgh$

This mechanical energy is lost in doingwork against friction on the flat part of the track.

So,

$\text{Loss in mechanical energy=work done against friction}$

$\Rightarrow mgh=\mu mgs$

Where, $s$ is the distance travelled by the particle before coming to rest.

$\Rightarrow s=\frac{h}{\mu }=\frac{1.5}{0.2}=7.5$



Let $E$ be the point where the particle comes to rest.

After starting from $B$ the particle will reach $C$ and will rise up till the remaining $K.E$ at $C$ is converted into potential energy.

It will then again descend and at $C$ will have the same value of mechanical energy as it had while ascending, but now it will move from $C$ to $B$. The same will be repeated and finally the particle will come to rest at $E$ such that

$BC+CB+BE=7.5$

$\Rightarrow 3+3+BE=7.5$

$\therefore BE=1.5\text{m}$

So the particle comes to rest at the center of the flat part.

Hence, option (b) is the correct answer.