Question

Under the action of a force, a $2kg$ body moves such that its position $x$ (in meters) as a function of time $t$ (in seconds) is given by $x=t^2/2$. The work done by the force in the first $5$ seconds is:

• A. 2.5 J
• B. 0.25 J
• C. 25 J
• D. 250 J

Answer 1

The correct option is C 25 J

Work Done $W=\int \overrightarrow{F}.d\overrightarrow{s}$

$x=\frac{t^2}{2}$

On differentiating, we will get

$\frac{dx}{dt}=t$

$\frac{d^{2}x}{d{t}^2}=1$

Force along the direction of motion $F=m{a}_x=m\frac{d^{2}x}{d{t}^2}=2\left(1\right)=2N$

$dx=tdt$
Thus for the current problem,
$W=\int Fdx={\int }_0^{5}2.tdt=2[\frac{t^2}{2}{]}_0^5=25J$