Question

Under the action of a force, a $2kg$ 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 meter and $t$ in second. The work done by the force in first two seconds is:

• A. $1600J$
• B. $160J$
• C. $16J$
• D. $\frac{16}{9}J$

Answer 1

The correct option is C $16J$

Given,

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

Differentiate it wrt t

$\frac{dx}{dt}=\frac{3}{3}{t}^2$

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

$v=t^2$

Workdone (W)= Change in K.E

$W=K.E_f-K.E_i$

$K.E_i=0$

$K.E_f=\frac{1}{2}m{v}^2$

$K.E_f=\frac{1}{2}\times 2\times 4^2$

$K.E_f=16J$

$W=16J$