Question

A force acts on a $30g$ particle in such a way that the position of the particle as a function of time is given, $x=3t-4t^2+t^3$. where $x$ is in meters and $t$ is in second. The work done during the first 4 seconds is:

• A. $5.28J$
• B. $450mJ$
• C. $490mJ$
• D. $530mJ$

Answer 1

The correct option is A $5.28J$

The mass is $30g$ or $0.03kg$ and the position of the particle is $x=3t-4t^2+t^3$.

The velocity is given as,

$\frac{dx}{dt}=3-8t+3t^2$

The acceleration is given as,

$\frac{d^{2}x}{d{t}^2}=-8+6t$

The work done is given as,

$dW=\left(ma\right)dx$

$dW=\left(0.03\right)\times \left(-8+6t\right)\left(3-8t+3t^2\right)dt$

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

$W=\left(0.03\right){\left(18\times \frac{t^4}{4}-72\times \frac{t^3}{3}+82\times \frac{t^2}{2}-24t\right)}_0^4$

$W=\left(0.03\right)\times 176$

$W=5.28J$

Thus, the work done during the first $4s$ is 5.28$J$.