# A body of mass 0.5 kg travels in a straight line with velocity

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

solution:

Mass of the body, m = 0.5 kg

Velocity of the body is governed by the equation, $v=a x^{\frac{3}{2}}$ with $a=5 \mathrm{~m}^{\frac{-1}{2}} \mathrm{~s}^{-1}$

Initial velocity, $u($ at $x=0)=0$

Final velocity $v($ at $x=2 \mathrm{~m})=10 \sqrt{2} \mathrm{~m} / \mathrm{s}$

Work done, $W=$ Change in kinetic energy

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

$=\frac{1}{2} \times 0.5\left[(10 \sqrt{2})^{2}-(0)^{2}\right]$

$=\frac{1}{2} \times 0.5 \times 10 \times 10 \times 2$

$=50 \mathrm{~J}$