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 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

The Correct Option is D

The correct option is(D): 530 mJ.

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