Work Done by a Spring Force — Question 1

Question 1

An endpoint force is not an average spring force

Assume ideal massless Hookean springs with fixed supports unless stated otherwise. A spring has $k=200$ N/m. An agent stretches it slowly from $x=0$ to $x=0.30$ m, where x is extension.

1. Find spring work.

2. Find the displacement-averaged spring force.

3. Explain the error in multiplying the final spring force by the displacement.

Question 1 - Solution

An endpoint force is not an average spring force

(a) Integrate the restoring force.

The governing relation is $$\boxed{F_s=-kx,\qquad W_s=\int_{x_i}^{x_f}(-kx)\,dx=\tfrac12k(x_i^2-x_f^2)}$$

$$W_s=\int_0^{.30}-200x\,dx=-100(.30)^2=-9\ \mathrm J.$$

(b) Average over displacement.

$$\overline F_s=W_s/.30=-30\ \mathrm N.$$ The slowly moving agent does $+9\ \mathrm J$ of work.

(c) Use the full force history.

The force changes linearly from 0 to $-60\ \mathrm N$. Using $$(-60)(.30)=-18\ \mathrm J$$ counts a rectangle instead of the triangular signed area, doubling the magnitude.