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For a particular nonlinear spring, the relationship between the magnitude of the applied force F and the resultant displacement x from equilibrium is given by the equation: F = Kx^2. What is the amount of work done by stretching the spring a distance xo?

Respuesta :

xero099

Answer:

[tex]F = \frac{k\cdot x_{o}^{3}}{3}[/tex]

Explanation:

Given that force changes with distance. From Physics we remember that work can be defined by the following integral equation:

[tex]W = \int\limits^{x_{b}}_{x_{a}} {F} \, dx[/tex] (1)

Where:

[tex]F[/tex] - Force.

[tex]W[/tex] - Work.

[tex]x[/tex] - Distance.

If we know that [tex]F = k\cdot x^{2}[/tex], then the work of the nonlinear spring is:

[tex]W = k\cdot \int\limits^{x_{b}}_{x_{a}} {x^{2}} \, dx[/tex]

[tex]W = k\cdot \left(\frac{x_{b}^{2}-x_{a}^{2}}{3} \right)[/tex]

[tex]W = \frac{k}{3}\cdot (x_{b}^{3}-x_{a}^{3})[/tex] (2)

If we know that [tex]x_{a} = 0[/tex] and [tex]x_{b} = x_{o}[/tex], then the amount of work done by stretching the spring a distance [tex]x_{o}[/tex] is:

[tex]F = \frac{k\cdot x_{o}^{3}}{3}[/tex]

onyebuchinnaji

The work done by stretching the spring to the given distance is [tex]W = \frac{kx_0}{3}[/tex].

The given parameters:

  • Applied force on the spring = F
  • Extension of the spring, = x₀

The work done by stretching the spring to the given distance is calculated as follows;

[tex]W = \int\limits^{x_b} _{x_a} {F} \, dx \\\\ W= \int\limits^{x_b} _{x_a} {kx^2} \, dx \\\\W = k\int\limits^{x_b} _{x_a} {x^2} \, dx\\\\ W= k[\frac{x^3}{3} ]\\\\W = k[\frac{x_b - x_a}{3} ]\\\\ W= k [\frac{x_0 - 0}{3} ]\\\\W = \frac{kx_0}{3}[/tex]

Thus, the work done by stretching the spring to the given distance is [tex]W = \frac{kx_0}{3}[/tex].

Learn more about work done here: https://brainly.com/question/8119756

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