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A cube of mass M starts at rest from point 1 at a height 4R, where R is the radius of track. The cube slides down the frictionless track and around the loop. The force which the track exerts on the cube at point 2 is:

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3 Mg

Cube will not reach the point 2

2 Mg

answer is A.

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Detailed Solution

Potential energy of cube at position 1, $U_{1} = M g (4 R) = 4 M g R$

At position 2, $U_{1} = U_{2} + K_{2}$

$4 M g R = 2 M g R + \frac{1}{2} M v^{2}$

$4 M g R = M v^{2}$

$v^{2} = 4 g R$

$N = \frac{M v^{2}}{R} - M g$

$N = \frac{M}{R} (4 R g) - M g = 3 M g$

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