Question

Figure shows a smooth track, a part of which is a circle of radius $r$. A block of mass $m$ is pushed against a spring of spring constant $k$ fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force $mg$ when it reaches the point $P$, where the radius of the track is horizontal.

• A. $\sqrt{\frac{6mgr}{k}}$
• B. $\sqrt{\frac{mgr}{k}}$
• C. $\sqrt{\frac{3mgr}{k}}$
• D. $\sqrt{\frac{2mgr}{k}}$

Answer 1

The correct option is C $\sqrt{\frac{3mgr}{k}}$

Let the initial compression of the spring be $x$.
Since the track is smooth, and normal reaction $N\perp v$ at all times, hence $N.ds=0\&W_N=0$
Also, gravity is a conservative force.


Applying equation of dynamics towards centre at point $P$,
$N=\frac{m{v}^2}{r}$
Given that at $P$, value of $N=mg$
$\Rightarrow mg=\frac{m{v}^2}{r}...\left(i\right)$

When spring is compressed by $x$, then elastic $PE$ stored, $P{E}_{\text{spring}}=\frac{1}{2}k{x}^2$
Gain in $KE=\frac{1}{2}m{v}^2$
Gain in $P{E}_{\text{grav}}=mgr$
$\therefore$ Applying mechanical energy conservation from initial position to point $P$:
$\text{Loss in stored}P{E}_{\text{spring}}=\text{Gain in}KE+\text{Gain in}P{E}_{\text{grav}}$
$\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2+mgr...\left(ii\right)$
From Eq. $\left(i\right)$ and $\left(ii\right)$
$\frac{1}{2}k{x}^2=\frac{1}{2}m\left(rg\right)+mgr$
$\Rightarrow k{x}^2=3mgr$
$\therefore x=\sqrt{\frac{3mgr}{k}}$