Question

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?

Answer 1

Answer:

$F = \frac{k\cdot x_{o}^{3}}{3}$

Explanation:

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

$W = \int\limits^{x_{b}}_{x_{a}} {F} \, dx$ (1)

Where:

$F$ - Force.

$W$ - Work.

$x$ - Distance.

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

$W = k\cdot \int\limits^{x_{b}}_{x_{a}} {x^{2}} \, dx$

$W = k\cdot \left(\frac{x_{b}^{2}-x_{a}^{2}}{3} \right)$

$W = \frac{k}{3}\cdot (x_{b}^{3}-x_{a}^{3})$ (2)

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

$F = \frac{k\cdot x_{o}^{3}}{3}$

Answer 2

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

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;

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

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

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