Work Done by a Spring Force — Question 2

Question 2

Equal compression and extension hide an intermediate speed maximum

Assume ideal massless Hookean springs with fixed supports unless stated otherwise. A 1 kg attached block moves on a frictionless horizontal rail. The spring has $k=100$ N/m. It passes $x=-0.20$ m moving right at 1 m/s.

1. Find spring work from $-0.20$ to $+0.20$ m.

2. Find speed at $x=0$ and at $+0.20$ m.

3. Locate the next turning point and explain why zero net work does not mean constant speed.

Question 2 - Solution

Equal compression and extension hide an intermediate speed maximum

(a) Compare squared extensions.

The governing relation is $$\boxed{K(x)=K_i+\tfrac12k(x_i^2-x^2)}$$

$$W_s=\tfrac12k(x_i^2-x_f^2)=0\ \mathrm J.$$

(b) Resolve the two parts of the trip.

The spring supplies 2 J before $x=0$, then removes 2 J. Initial $$K=0.5\ \mathrm J;$$ at zero $$K=2.5\ \mathrm J,\qquad v=\sqrt5\ \mathrm{m/s}.$$ At $+0.20\ \mathrm m$, speed is again 1 m/s.

(c) Find where the remaining kinetic energy vanishes.

$$0=0.5+50(.04-x_t^2)$$ gives $$x_t=\sqrt{.05}=0.2236\ \mathrm m.$$ Speed is largest at zero because spring work changes from positive to negative there.