Question

Numerical integration. A breadbox is made to move along an x axis from $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{m}$ by a force with a magnitude given by $\mathbf{F}\mathbf{=}\mathbf{exp}\mathbf{(}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{)}$, with x in meters and F in Newton. (Here exp is the exponential function.) How much work is done on the breadbox by the force?

Short answer

Work done by the force F on the breadbox is $W=0.47\mathrm{J}$.

Step-by-step solution

Given information

It is given that,

$$\begin{gathered} {\mathrm{x}}_1=0.15\mathrm{m} \\ {\mathrm{x}}_2=1.20\mathrm{m} \\ \mathrm{F}=\exp \left(-2{\mathrm{x}}^2\right) \end{gathered}$$

Determining the concept

The problem deals with the work done which is the fundamental concept of physics.Work is the displacement of an object when force is applied on it.Use the concept of work done. Force as the function of x, and the initial and final positions are given. Integrate the force over given limits to find the work done.

Formula:

W=F.d

where,F is force, dis displacement and Wis the work done.

Determining thework done by the force F on the breadbox

Use integration to find the work done by force,

$$\begin{gathered} W=\int F.dx \\ ={\int }_{0.15}^{1.20}{e}^{\left(-2x^2\right)}dx \\ W={\left[\frac{1}{2}\sqrt{\frac{\pi }{2}}erf\left(\sqrt{2x}\right)\right]}_{0.15}^{1.20} \end{gathered}$$

Putting the limits values,

$W=0.47\mathrm{J}$

Hence,the work done by theforce F on the breadbox is $W=0.47\mathrm{J}$.