Question

Under the action of a force, a $2\text{kg}$ body moves such that its position $x$ as a function of time $t$ is given by $x=\frac{t^3}{3}$, where $x$ is in $\text{meter}$ and $t$ in $\text{second}$. The work done by the force in the first two seconds is (Give an integer value)

Answer 1

Given,

$x=\frac{t^3}{3}$

Now, by differentiating the above equation, we get the velocity at ant time $t$ as

$v=\frac{dx}{dt}=t^2$

At $t=0,$ $v=0$ and

At $t=2\text{s}$, $v=4\text{m/s}$

From work energy theorem,

$\text{Work one by the force(W)=change in kinetic energy}$

$W=K.E_f-K.E_i$

$\Rightarrow W=\frac{1}{2}M(v_f^2-v_i^2)$

$\Rightarrow W=\frac{1}{2}\times 2\times (4^2-0^2)$

$\therefore W=16\text{J}$

Accepted answer : $16$