Question

A force acts on a $3g$ 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 metres and $t$ is in seconds. The work done during the first $4$ second is

• A. 490 mJ
• B. 450 mJ
• C. 576 mJ
• D. 530 mJ

Answer 1

Answer: (D)

We have,
mass, $m=3g=0.003kg$
$x=3t-4t^2+t^3$
Now, $v=\frac{dx}{dt}=3-8t+3t^2\Rightarrow dx=\left(3-8t+3t^2\right)dt$
$\Rightarrow a=\frac{dv}{dt}=0-8+6t$
Now, $dw=Fdx$
$\Rightarrow dw=(ma)dx$
$\Rightarrow dw=(0.003)(-8+6t)\left(3-8t+3t^2\right)dt$
$\Rightarrow dw=(0.003)\left(18t^3-72t^2+82t-24\right)dt$
$\Rightarrow w=(0.003){\int }_0^4\left(18t^3-72t^2+82t-24\right)dt$
$\Rightarrow w=0.003\times 176=0.528J$
$\Rightarrow w=530mJ$