Question

A cube of mass M starts at rest from point 1 at a height 4R, where R is the radius of the circular 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:

Answer 1

At point 2, we can find velocity by energy conservation
\(Mg.4R = mg.2R + \frac{1}{2}mv^2 \\
\Rightarrow 2mgR = \frac{1}{2}mv^2 \Rightarrow V=\sqrt{4gR}\)
Now,
$N=\frac{m{v}^2}{R}-mg=\frac{m.4gR}{R}-mg$
=3 mg