Question

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

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

Answer 1

Answer: (A)

20 J

Answer 2

Kinetic energy is the energy possessed by a body of mass m due to its velocity $v$.
$\therefore KE = \frac{1}{2} mv^2$
At the highest point velocity, $v_A = \sqrt{rg}$
$KE = \frac{1}{2} mv_A^2$
At the lowest point velocity, $v_B = \sqrt{ 5rg }$
$KE = \frac{1}{2} mv_B^2$
Difference in $KE, \Delta K = \frac{1}{2} m ( v_B^2 - v_A^2 )$
$= \frac{1}{2} m ( 5 rg - rg )$
$= 2m (rg )$
$= 2 \times 1 \times 1 \times 1 \times 10$
$= 20\, J$