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