To find the work done by the force in moving the particle from [tex]\( x = 0 \)[/tex] to [tex]\( x = 2x_0 \)[/tex], we need to integrate the force function over the given displacement.
1. Understand the Force Function:
The force [tex]\( F \)[/tex] is given by:
[tex]\[
F(x) = F_0 \left( \frac{x}{x_0} - 1 \right)
\][/tex]
Here, [tex]\( F_0 = 0.71 \, \text{N} \)[/tex] and [tex]\( x_0 = 2.5 \, \text{m} \)[/tex].
2. Set up the Integral for Work:
Work done by a force over a displacement is given by the integral of the force over that displacement. Thus, the work [tex]\( W \)[/tex] is:
[tex]\[
W = \int_{x_{\text{start}}}^{x_{\text{end}}} F(x) \, dx
\][/tex]
Given [tex]\( x_{\text{start}} = 0 \)[/tex] and [tex]\( x_{\text{end}} = 2x_0 \)[/tex], we have:
[tex]\[
W = \int_{0}^{2x_0} F_0 \left( \frac{x}{x_0} - 1 \right) dx
\][/tex]
Substituting [tex]\( F_0 \)[/tex] and [tex]\( x_0 \)[/tex] into the equation, we get:
[tex]\[
W = \int_{0}^{2 \cdot 2.5} 0.71 \left( \frac{x}{2.5} - 1 \right) dx
\][/tex]
3. Simplify the Integral:
Simplify the integrand:
[tex]\[
W = 0.71 \int_{0}^{5} \left( \frac{x}{2.5} - 1 \right) dx
\][/tex]
[tex]\[
= 0.71 \int_{0}^{5} \left( \frac{x}{2.5} - 1 \right) dx
\][/tex]
[tex]\[
= 0.71 \left( \int_{0}^{5} \frac{x}{2.5} \, dx - \int_{0}^{5} 1 \, dx \right)
\][/tex]
4. Evaluate the Integrals:
Evaluate each integral individually:
[tex]\[
\int_{0}^{5} \frac{x}{2.5} \, dx = \frac{1}{2.5} \int_{0}^{5} x \, dx = \frac{1}{2.5} \left[ \frac{x^2}{2} \right]_{0}^{5} = \frac{1}{2.5} \left( \frac{25}{2} - 0 \right) = \frac{25}{5} = 5
\][/tex]
[tex]\[
\int_{0}^{5} 1 \, dx = x \big|_{0}^{5} = 5 - 0 = 5
\][/tex]
5. Combine Results:
Substituting these results back into the expression for [tex]\( W \)[/tex]:
[tex]\[
W = 0.71 \left( 5 - 5 \right) = 0.71 \cdot 0 = 0
\][/tex]
Thus, the work done by the force in moving the particle from [tex]\( x = 0 \)[/tex] to [tex]\( x = 2x_0 \)[/tex] is extremely close to [tex]\( 0 \)[/tex] (numerically, it is [tex]\( 7.073458514357402 \times 10^{-17} \)[/tex] Joules), essentially confirming that the net work done is negligible (effectively zero in practical terms).
