Q. 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 :
Delhi UMET/DPMTDelhi UMET/DPMT 2005Work, Energy and Power
Solution:
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 $
$= 20\, J$
