Question

A force acts on a $30gm$ 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 on the particle during the first $4$ second is

• A. $3.84J$
• B. $1.68J$
• C. $5.28J$
• D. $5.41J$

Answer 1

The correct option is C $5.28J$

Given,

$m=30g=\frac{30}{1000}=0.03kg$

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

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

Velocity, $v=\frac{dx}{dt}$

$v=3-8t+3t^2$

Acceleration, $a=\frac{dv}{dt}$

$a=-8+6t$

Work done on the particle during the first $4$ second

$W=\int F.dx=m\int adx$

$W=m{\int }_0^4(-8+6t)(3-8t+3t^2)dt$

$W=0.03{\int }_0^4(18t^3-72t^2+82t-24)dt$

$W=0.03[\frac{18}{4}{t}^4-\frac{72}{3}{t}^3+\frac{82}{2}{t}^2-24t{]}_0^4$

$W=0.03\times 176$

$W=5.28J$

The correct option is C.