Question

Displacement, $x$ (in meters), of a body of mass $1kg$ 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
• E. 16 J

Answer 1

Answer: (D)

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