Physics - W.E.P. Question with Solution | TestHub

For the block to just complete the loop, its speed at the highest point must be $v_{top}=\sqrt{gR}$. By conservation of mechanical energy from the initial height $H$ to the highest point of the loop (height $2R$), $mgH=mg(2R)+\frac{1}{2}m{v}_{top}^2$. Substituting $v_{top}^2=gR$, we get $mgH=2mgR+\frac{1}{2}m(gR)$, so $H=2.5R$. Now, by conservation of energy from height $H$ to the lowest point of the loop (height 0), $mgH=\frac{1}{2}m{v}_{bottom}^2$. Substituting $H=2.5R$, we get $mg(2.5R)=\frac{1}{2}m{v}_{bottom}^2$, so $v_{bottom}^2=5gR$. At the lowest point, the normal force $N_{bottom}$ and gravity $mg$ provide the centripetal force: $N_{bottom}-mg=\frac{m{v}_{bottom}^2}{R}$. Substituting $v_{bottom}^2=5gR$, we get $N_{bottom}-mg=\frac{m(5gR)}{R}=5mg$. Therefore, $N_{bottom}=6mg$. The normal force is 6 units of $mg$.