Question

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

• A. $1.5J$
• B. $50J$
• C. $10J$
• D. $100J$

Answer 1

The correct option is B $50J$

Given: $m=0.5kg,v=k{x}^{3/2}$ where, $k=5m^{-1/2}{s}^{-1}$
Acceleration, $a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}(\because v=\frac{dx}{dt})$
As $v^2=k^2{x}^3$
Diferentiating both sides with respect to $x$, we get
$2v\frac{dv}{dx}=3k^2{x}^2$
$\therefore Acceleration,a=\frac{3}{2}{k}^2{x}^2$
Force, $F=Mass\times Acceleration=\frac{3}{2}m{k}^2{x}^2$
Work done, $W=\int Fdx={\int }_0^2\frac{3}{2}m{k}^2{x}^{2}dx$
$W=\frac{3}{2}m{k}^2{\left[\frac{x^3}{3}\right]}_0^2=\frac{3}{6}\times 0.5\times 5^2\times [2^3-0]=50J$.