Question

A particle of mass $M$ rests on a straight groove along which it is constrained to move. A perfectly elastic rubber band of natural length $l$ and uniform area of cross-section is attached with the particle. The other end of the band is suspended from a rigid support. A force $K(l^{\prime 2}-l^2{)}^{1/2}$ is required to stretch the band to a length $l^{\prime }$. The particle is moved to a distance $S$ (where $S<<$ $l$) and then released. Taking $K=\frac{Mg}{S}$ and $\mu$ as the coefficient of friction between the particle and the groove, the velocity of particle when passing through the initial position is:

• A. ${\left[\frac{gS}{3l}\right(2S-3\mu l\left)\right]}^{1/2}$
• B. $\left[\frac{gS}{3l}\right(3S-3\mu l{)}^{1/2}]$
• C. $\frac{gS}{l}(3S-2\mu l{)}^{1/2}$
• D. ${\left[\frac{gS}{2l}\right(3S-2\mu l\left)\right]}^{1/2}$

Answer 1

The correct option is A ${\left[\frac{gS}{3l}\right(2S-3\mu l\left)\right]}^{1/2}$

Let the particle be at a distance $x$ at any instant and $T$ be the tension in the string then
$T=K(l^{\prime 2}-l^2{)}^{1/2}=Kx$
Net force tending to make the particle move further through $dx$,
Work done $=[\frac{K{x}^2}{l}-\mu (Mg-Kx\left)\right]dx$

$\frac{1}{2}M{v}^2={\int }_0^S[\frac{K{x}^2}{l}-\mu (Mg-Kx\left)\right]dx$

$={[\frac{K{x}^3}{3l}-\mu (Mgx-\frac{K{x}^2}{2}\left)\right]}_0^S$

Substituting $K=\frac{Mg}{S}$

$={[\frac{Mg{x}^3}{3Sl}-\mu (Mgx-\frac{Mg{x}^2}{2S}\left)\right]}_0^S$

$=[\frac{Mg{S}^3}{3lS}-\mu (MgS-\frac{Mg{S}^2}{2S}\left)\right]$

$=[\frac{Mg{S}^2}{3l}-\mu (MgS-\frac{MgS}{2}\left)\right]$

$=[\frac{Mg{S}^2}{3l}-\mu (\frac{MgS}{2}\left)\right]$

$\frac{1}{2}M{V}^2=\frac{MgS}{6l}(S-\frac{3}{2}\mu l)$

or $v={\left[\frac{gS}{3l}\right(2S-3\mu l\left)\right]}^{1/2}$