Question

Under the action of a force a $2 \, kg$ body moves such that its position $x$ in meters as a function of time $t$ is given by $x=\frac{t^{4}}{4}+3$ . Then work done by the force in first two seconds is

• A. $6 \, J$
• B. $10 \, J$
• C. $7 \, J$
• D. $64 \, J$

Answer 1

Answer: (D)

$x=\frac{t^{4}}{4}+3$
$\frac{d x}{d t}=\frac{4 t^{3}}{4}+0$
$v=t^{3}$
Using work energy theorem
$w=\Delta k=\frac{1}{2}m\left(v \left(2\right)\right)^{2}-\frac{1}{2}m\left(v \left(0\right)\right)^{2}$
$=\frac{1}{2}\left(2\right)\left[\left(2^{3}\right)^{2} - 0^{3}\right] \, $
$=64 \, J$