Question

The force acting on a particle is $Fx=F_0{e}^{-x/L}$ . How much work does this force do as the particle moves along the x-axis from $x=0$ to $x=L$?

Short answer

The work done on the particle as it moves from$x=0tox=L$is$F_{0}L(1-\frac{1}{e}).$

Step-by-step solution

Content Introduction

The work done by the force $F_x$on the particle if the movement of particle is in same direction as of the force can be expressed as:

$W={\int }_{x_i}^{x_f}{F}_{x}dx$

Here, $F_x$is the force and $dx$is the small displacement in the direction of the force.

Content Explanation

The work done on the particle as it moves from $x=0tox=L$is shown as follow:

role="math" localid="1647794367778" $$\begin{gathered} W={\int }_0^L{F}_{x}dx \\ W={\int }_0^L{F}_0{e}^{-x/L}dx \\ W=F_0{{(-L{e}^{-x/L})}^L}_{0}W=F_{0}L(1-\frac{1}{e}) \end{gathered}$$