Question

Displacement, $x$ (in meters), of a body of mass $1\,kg$ as a function of time, $t$, on a horizontal smooth surface is give as $x = 2t^2$. The work done in the first one second by the external force is

• A. 1 J
• B. 2 J
• C. 4 J
• D. 8 J

Answer 1

The Correct Option is D

Given displacement is,
$X =2 t^{2}$
$\Rightarrow V =\text { velocity }=\frac{d x}{d t}=4 t $
$V_{\text {inital }} =V(t=0) $
$=4 \times 0=0 \,m / s $
$V_{\text {final }} =V(t=1) $
$=4 \times 1=4\, m / s$
$\Delta K . E =$ change in $K . E$ of body
$=\frac{1}{2} m\left(V_{\text {final }}^{2}-V_{\text {initial }}^{2}\right) $
$=\frac{1}{2} \times 1 \times(16-0)=8 \,J$
By work-kinetic energy theorem, Work done $=\Delta K$. $E$
$=8\, J$