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How much work does this force do as the particle moves along the x-axis from x = 0 to x = l? express your answer in terms of the variables f0 and l?

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