Work Done by a Spring Force — Question 3

Question 3

The coordinate origin need not be the relaxed position

Assume ideal massless Hookean springs with fixed supports unless stated otherwise. A spring acts on a slider along laboratory coordinate q. It is relaxed at $q=0.40$ m and has $k=50$ N/m. The slider moves from $q=0.10$ to $0.60$ m.

1. Write the spring force as a function of q.

2. Find spring work between these endpoints.

3. Explain why substituting q directly into $\tfrac12k(q_i^2-q_f^2)$ fails.

Question 3 - Solution

The coordinate origin need not be the relaxed position

(a) Subtract the relaxed coordinate.

The governing relation is $$\boxed{x=q-q_0,\qquad W_s=-k\int_{q_i}^{q_f}(q-q_0)\,dq}$$ $x=q-0.40$, so $$F_s=-50(q-0.40)\ \mathrm N$$ when q is in meters.

(b) Use actual deformations.

$$x_i=-0.30\ \mathrm m,\qquad x_f=+0.20\ \mathrm m.$$ Thus $$W_s=25(.09-.04)=1.25\ \mathrm J.$$

(c) Check the force zero.

The formula using unshifted q would give $-8.75\ \mathrm J$ and assumes the spring is relaxed at q=0. A coordinate shift is harmless only if the force law is shifted with it.