Question

A force acts on a $3$ gm 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 $4$ second is:

• A. $384mJ$
• B. $168mJ$
• C. $528mJ$
• D. $541mJ$

Answer 1

The correct option is C $528mJ$

Mass of the particle,
$m=0.003kg$

$x=3t-4t^2+t^3$

$\therefore a=-8+6t$

Thus force acting on the particle is $ma$

Thus the work done on the particle$=\int Fdx$
$={\int }_0^{4}m(-8+6t)(3-8t+3t^2)dt$

$=528mJ$ -m represents mass here.