Question

A body of mass $1kg$ is moving in a vertical circular path of radius $1m$. The difference between the kinetic energies at the highest and lowest point is :

• A. $20J$
• B. $10J$
• C. $4\sqrt{5}J$
• D. $10(\sqrt{5}-1)J$

Answer 1

The correct option is A $20J$

Given,

$r=1m$

$m=1kg$

$g=10m/s^2$

In limiting case, for body to revolve in circle,

$V_{top}=v_1=\sqrt{rg}$

$V_{bott}=v_2=\sqrt{5rg}$

The difference in kinetic energy,

$\Delta E=(K.E{)}_{bott}-(K.E{)}_{top}$

$\Delta =\frac{1}{2}m(v_2^2-v_1^2)$

$\Delta =\frac{1}{2}m(5rg-rg)=\frac{4mrg}{2}$

$\Delta E=\frac{4\times 1\times 1\times 10}{2}$

$\Delta E=20J$

The correct option is A.