Question

A force acts on a $3$g particle in such a way that the position of the particle as a function of time is given by $x=3t-4t^2+t^3$, where $x$ is in meters and $t$ is in seconds. The work done during the first $4$ second is :

• A. $2.88$ J
• B. $450$ mJ
• C. $490$ mJ
• D. $530$ mJ

Answer 1

The correct option is D $530$ mJ

We have,

mass, $m=3g=0.003kg$

$x=3t-4t^2+t^3$

Now,

$v=\frac{dx}{dt}=3-8t+3t^2\Rightarrow dx=(3-8t+3t^2)dt$

$\Rightarrow a=\frac{dv}{dt}=0-8+6t$

Now,

$dw=Fdx$

$\Rightarrow dw=\left(ma\right)dx$

$\Rightarrow dw=\left(0.003\right)(-8+6t)(3-8t+3t^2)dt$

$\Rightarrow dw=\left(0.003\right)(18t^3-72t^2+82t-24)dt$

$\Rightarrow w=\left(0.003\right){\int }_0^4(18t^3-72t^2+82t-24)dt$

$\Rightarrow w=0.003\times 176=0.528J$

$\Rightarrow w=530mJ$