Respuesta :
Answer;
= F0 L ( 1 - 1/e )
Explanation;
Work done is given as the product of force and distance.
In this case;
Work done = ∫︎ F(x) dx
= F0 ∫︎ e^(-x/L) dx
= F0 [ -L e^(-x/L) ] between 0 and L
= F0 L ( 1 - 1/e )
The work required for this force to do as the particle moves along the x-axis from x = 0 to x = l is,
[tex]W=F_oL(1-\dfrac{1}{e})[/tex]
What is work done?
Work done is the force applied on a body to move it over a distance.
The work done by a fixed force on a point that displaces the particle in the direction of force can be given as,
[tex]W=\int {F(x)} \, dx[/tex]
Here, (F) is the magnitude of the force, and (x) is the distance moved by the particle.
Now, the displacement of the particle is along the x-axis from x = 0 to x = l. Thus, the limit of the integration should be 0 to I as,
[tex]W=F_0\int\limit^I_0 {(e^{(-x/l)})} \, dx[/tex]
Integrate the above function, as,
[tex]W=F_0\left[-le^\dfrac{-x}{l} \right]^1_0\\W=F_oL(1-\dfrac{1}{e})[/tex]
Thus, the work required for this force to do as the particle moves along the x-axis from x = 0 to x = l is,
[tex]W=F_oL(1-\dfrac{1}{e})[/tex]
Learn more about the work done here;
https://brainly.com/question/25573309