Question

A single force acts on a 3.0 kgparticle-like object whose position is given by$\mathbf{x}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{t}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{0}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{0}{\mathbf{t}}^{\mathbf{3}}$, with x in meters and t in seconds. Find the work done by the force from$\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{to}\mathbf{}\mathbf{t}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{s}$.

Short answer

The work done by the force from $\mathrm{t}=0\mathrm{to}\mathrm{t}=4.0\mathrm{s}$is$5.28\times {10}^2\mathrm{J}$

Step-by-step solution

Given

1. The mass of the particle$\mathrm{m}=3.0\mathrm{kg}$
2. The position of the particle$\mathrm{x}=3.0\mathrm{t}-4.0{\mathrm{t}}^2+1.0{\mathrm{t}}^3$

Understanding concept of work done

The problem deals with the concept of work done. It includes both forces exerted on the body and the total displacement of the body. We can find the work done by finding the work done by the given force over a small displacement and integrate it over the given limits.

Formula:

$$\begin{gathered} \mathbf{W}\mathbf{=}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}\mathbf{F}\left(x\right)\mathbf{d}\mathbf{x} \\ \mathbf{v}\mathbf{=}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}} \end{gathered}$$

Calculate the work done

The work done by a force is given by

$\mathrm{W}=\int \mathrm{Fdx}$

But, we can also write,

$\begin{array}{rcl}\mathrm{F}& =& \mathrm{ma}\\ & =& \mathrm{m}\frac{d\mathrm{v}}{d\mathrm{t}}\end{array}$

So,

$\begin{array}{rcl}\mathrm{W}& =& \int \mathrm{F}\frac{d\mathrm{v}}{d\mathrm{t}}\mathrm{dx}\\ & =& \int \mathrm{m}\frac{d\mathrm{v}}{d\mathrm{t}}\mathrm{dx}\\ & =& \int \mathrm{m}\frac{d\mathrm{x}}{d\mathrm{t}}\mathrm{dv}\\ & =& \int \mathrm{mvdv}\end{array}$

We are given the expression for x in terms of t. By differentiating the expression with respect to t, we can determine the expressions for v and dv.

$\begin{array}{rcl}\mathrm{x}& =& 3.0\mathrm{t}-4.0{\mathrm{t}}^2+1.0{\mathrm{t}}^3\\ \mathrm{v}& =& \frac{d\mathrm{x}}{d\mathrm{t}}\\ & =& 3.0-8.0\mathrm{t}+3.0{\mathrm{t}}^2\end{array}$

And,

$\mathrm{dv}=\left(-8.0+6.0\mathrm{t}\right)\mathrm{dt}$

Now,

$$\begin{gathered} \mathrm{vdv}=\left(3.0-8.0\mathrm{t}+3.0{\mathrm{t}}^2\right)\left(-8.0+6.0\mathrm{t}\right)\mathrm{dt} \\ \mathrm{vdv}=\left(-24+82\mathrm{t}-72{\mathrm{t}}^{2}18{\mathrm{t}}^3\right)\mathrm{dt} \end{gathered}$$

Now, we can determine the work done as

$\begin{array}{rcl}\mathrm{W}& =& \int \mathrm{mvdv}\\ & =& {\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\left(3.0\right)\times \left(-24+82\mathrm{t}-72{\mathrm{t}}^2+18{\mathrm{t}}^3\right)d\mathrm{t}\\ \mathrm{W}& =& 3{\left[-24\mathrm{t}+82\frac{{\mathrm{t}}^2}{2}-72\frac{{\mathrm{t}}^3}{3}+18\frac{{\mathrm{t}}^4}{4}\right]}_0^4\\ \mathrm{W}& =& 5.28\times {10}^2\mathrm{J}\end{array}$

Therefore the work done by the force from $\mathrm{t}=0\mathrm{to}\mathrm{t}=4.0\mathrm{s}$is $5.28\times {10}^2\mathrm{J}$.