Question

A particle of mass $0.5kg$ travels in a straight line with velocity $v=a{x}^{3/2}$, where $a=5m^{-1/2}{s}^{-1}$. What is the work done by the net force during its displacement from $x=0$ to $x=2m$?

Answer 1

Mass $m=0.5kg$

Velocity ,$v=a{x}^{3/2}$

$a=5m^{-1/2}{s}^{-1}$

Initial velocity at $x=0=0$

Final velocity at $x=2=10\sqrt{2}m.s$

Work done =change in K.E

$=\frac{1}{2}m(v^2-u^2)$

$=\frac{1}{2}\times 0.5\times \left[200\right]=50J$

$WD=50J$