Question

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

• A. $2J$
• B. $4J$
• C. $8J$
• D. $16J$

Answer 1

Answer: (C)

Given displacement is,
$x=2t^2$
$\Rightarrow v=velocity=\frac{dx}{dt}=4t$
$v_{initial}=v\left(t=0\right)$
$=4\times 0=0m{s}^{-1}$
$v_{final}=v\left(t=1\right)$
$=4\times 1=4m{s}^{-1}$
$\Delta K.E=$ change in $K.E$ of body
$=\frac{1}{2}m\left(v_{final}^2-v_{initial}^2\right)$
$=\frac{1}{2}\times 1\times \left(16-0\right)=8J$
By work $-$ kinetic energy theorem, work done $=\Delta K.E=8J$