Question

Displacement $x$ (in meters), of a body of mass $1 \, kg$ 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. $2 \, J$
• B. $4 \, J$
• C. $8 \, J$
• D. $16 \, J$

Answer 1

Answer: (C)

Given displacement is,
$x=2t^{2}$
$\Rightarrow v=velocity=\frac{d x}{d t}=4t$
$v_{i n i t i a l}=v \, \left(t = 0\right)$
$= \, 4\times 0=0 \, ms^{- 1}$
$v_{f i n a l}=v \, \left(t = 1\right)$
$= \, 4\times 1=4ms^{- 1}$
$\Delta K.E=$ change in $ \, K.E$ of body
$=\frac{1}{2}m\left(v_{f i n a l}^{2} - v_{i n i t i a l}^{2}\right)$
$=\frac{1}{2}\times 1\times \left(16 - 0\right)=8J$
By work $-$ kinetic energy theorem, work done $=\Delta K.E=8J$