Question

A force acts on a $30\text{g}$ 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 meters and $t$ is in seconds. The work done during the first $4$ seconds is

• A. $5.28\text{J}$
• B. $450\text{mJ}$
• C. $490\text{mJ}$
• D. $530\text{mJ}$

Answer 1

The correct option is A $5.28\text{J}$

Velocity as a function of time is $v=\frac{dx}{dt}=3-8t+3t^2$
According to work energy theorem
$W=\frac{1}{2}m(v_4^2-v_0^2)$
$v_4=3-8\times 4+3\times 4^2=19{\text{ms}}^{-1}$
$v_0=3-0-0=3{\text{ms}}^{-1}$

$W=\frac{1}{2}\times 0.03\times ({19}^2-3^2)=5.28\text{J}$