Question

The displacement x (in m), of a particle of mass m (in Kg) is related to time t (in sec) by $t=\sqrt{x}+3$. Find the work done in the first six seconds. (in mJ)

Answer 1

Solution : We have,

$t=\sqrt{x}+3$

$\Rightarrow t-3=\sqrt{x}$

$\Rightarrow x=(t-3{)}^2$

we know,

$V=\frac{dx}{dt}=2(t-3)$

At $t=0_s,$

$V\left(0\right)=2(0-3)=-6m/s$

At $t=6_s$,

$V\left(b\right)=2(6-3)=6m/s$

$△KE=\frac{1}{2}m\left[V\right(6{)}^2-V\left(0{)}^2\right]$

$=\frac{1}{2}m\left[\right(6{)}^2-(-6{)}^2]$

$=0$

By work energy theorem,

$W=△KE=0.$