Question

A force acts on a $30 \, g$ 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. $5.28 \, J$
• B. $450 \, mJ$
• C. $490 \, mJ$
• D. $530 \, mJ$

Answer 1

Answer: (A)

$v=\frac{d x}{d t}=3-8t+3t^{2}$
Speed at $t=0$ is
$v_{0}=3 \, m / s$
Speed at $t=4$ is
$v_{4}=19 \, m / s$
According to work energy theorem
$W=\frac{1}{2}m\left(v_{4}^{2} - v_{0}^{2}\right)$
$W=\frac{1}{2}\times 0.03\times \left(1 9^{2} - 3^{2}\right)=5.28 \, J$