Question

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

• A. 570 mJ
• B. 450 mJ
• C. 490 mJ
• D. 528 mJ

Answer 1

The correct option is D 528 mJ

$\frac{dx}{dt}=3-8t+3t^2$

$dx=(3-8t+3t^2)dt$

$\frac{d^2}{d{x}^2}=a=-8+6t$

Force $F=ma=m(-8+6t)$

Work done $F.dx$

$F.(3-8t+3t^2)dt$

${\int }_o^{w}dw={\int }_0^{4}m(-8+6t)(3-8t+3t^2)dtw=m{\int }_0^4(-24+82t-72t^2+18t^3)dt=m\stackrel{4}{\underset{0}{[-24t+41t^2-24t^3+\frac{3}{2}{t}^4]}}=m\left(176\right)=3\times {10}^{-3}\times 176=528mJ$