JEE Main · PhysicsHard

$A block of mass m is pushed against a spring of constant k by a distance x and released. The block slides along a frictionless surface and then up a circular track of radius R . What is the minimum compression x required for the block to reach the top of the track without losing contact?$

Show correct answer & step-by-step solution

Correct answer: A — $x = \frac{5 m g R}{k}$

Solution

$At the top of the loop, the minimum speed v satisfies m v^{2} / R = m g, so v^{2} = g R .$
$By conservation of energy, \frac{1}{2} k x^{2} = \frac{1}{2} m v^{2} + m g (2 R) .$
$Substituting v^{2} = g R, we get \frac{1}{2} k x^{2} = \frac{1}{2} m g R + 2 m g R = \frac{5}{2} m g R .$
$Solving for x, we find x = \frac{5 m g R}{k} .$
$Hence the answer is (A).$