Q4.3 Work as an integral (S)

A certain type of spring generates a restoring force of the form F=-λx3, where x is the change in length of the spring. For a distortion of 1 cm, the magnitude of the restoring force is 2 N.
  1. What is the value of λ?
  2. How much work is done by the restoring force when the spring is compressed a distance 2 cm?
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Hint

Remember the force changes as the spring is compressed so you will need to integrate the force to get the work.
 

Solution

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Solution

In this problem, the force varies with the displacement. Therefore to find the work, we will have to integrate the force along the path. We must assume that both F and x are parallel, so that the problem is one dimensional. i.e. $\vec{F}\cdot d\vec{r} = F(x)dx$ (see Equation 3.2).
  1. If F=-2 N when x=0.01 m, then λ=-F/x3=2/(0.01)3 = 2×106 Nm-3.
  2. The work done by the restoring force is then
    \[ W = - \int_0^{0.02} \lambda x^3 dx = -\left[ \lambda \frac{x^4}{4}\right]^{0.02}_0 = - 0.08\,J \]