Question

A small block of mass $m$ slides along a smooth circular track of radius $R$ as shown in the figure. If it starts from rest at P, what is the resultant force acting on it at Q?

• A. $F=\sqrt{35}mg$
• B. $F=\sqrt{55}mg$
• C. $F=\sqrt{65}mg$
• D. $F=\sqrt{84}mg$

Answer 1

Answer: (C)

$F=\sqrt{65}mg$

Let the velocity of the block at point Q be $v$.

By Work-energy theorem : $W=\Delta K.E.$

$\therefore$ $mg(5R-R)=\frac{1}{2}m{v}^2-0$

$⟹m{v}^2=8mg$

Using circular motion equation : $N=\frac{m{v}^2}{R}$

$\therefore$ $N=\frac{8mgR}{R}=8mg$

Total force on the block at Q,

$F=\sqrt{N^2+mg}=\sqrt{(8mg{)}^2+(mg{)}^2}=\sqrt{65}mg$