Question

A force acts on a $20g$ 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 $4sec$ is:

• A. $-1.6J$
• B. $-1600J$
• C. $2.6J$
• D. $1600J$

Answer 1

The correct option is C $2.6J$

Mass $\left(m\right)=20gm=0.020kg$

$x=3t-4t^2+t^{3}m$

$\Rightarrow v=\frac{dx}{dt}=3-8t+3t^{2}m/s$

$\Rightarrow a=\frac{dv}{dt}=-8+6tm/s^2$

Force $=ma=0.120t-0.160N$

Work done $=dw=Fdx=F\times \left(\frac{dx}{dt}\right)\times dt$

$\Rightarrow dw=(0.120+0.160)\times (3-8t+3t^2)dt$

$=0.360t^3-1.440t^2+1.640t-0.450$

$\Rightarrow w=integralofdwfromt=0to4sec$

$=[0.090t^4-0.490t^3+0.82t^2-0.450t],t=0to4s$

$=2.88J.$

Hence, the answer is $2.88J.$