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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. The initial compression of the spring so that the block presses the track with a force $m g$ when it reaches the point P, where the radius of the track is horizontal is $\sqrt{\frac{p m g R}{k}} .$ Find the value of $p$ ?

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answer is 3.

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Detailed Solution

Applying equation of dynamics towards centre at point P, $N = \frac{m v^{2}}{r}$

Given that at P, value of $N = m g \Rightarrow m g = \frac{m v^{2}}{r} ..... (i)$

$∴$ Applying mechanical energy conservation from initial position to point P Loss in stored

${PE}_{spring} {= Gain in KE + Gain in PE}_{grav}$

$\frac{1}{2} k x^{2} = \frac{1}{2} m v^{2} + m g r ..... (i i)$

From Eq.(i) and (ii), $\frac{1}{2} k x^{2} = \frac{1}{2} m (r g) + m g r \Rightarrow k x^{2} = 3 m g r$
$∴ x = \sqrt{\frac{3 m g r}{k}}$

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