Work Done by a Spring Force — Question 6

Question 6

A vertical release passes equilibrium before it turns

Assume ideal massless Hookean springs with fixed supports unless stated otherwise. A 1 kg mass is attached to a vertical spring of $k=100$ N/m and released from rest when the spring is unstretched. Let x be downward extension and use $g=10$ m/s$^2$.

1. Write kinetic energy as a function of x on the first descent.

2. Find the first nonzero turning extension.

3. Find the equilibrium extension and the speed there.

Question 6 - Solution

A vertical release passes equilibrium before it turns

(a) Add gravity and spring work.

The governing relation is $$\boxed{K(x)=W_g+W_s=mgx-\tfrac12kx^2\geq0}$$

$$K(x)=mgx-\tfrac12kx^2=10x-50x^2\ \mathrm J,$$ with x in meters.

(b) Set kinetic energy to zero.

$$x(10-50x)=0$$ gives the lower turning point $$x=0.20\ \mathrm m.$$ At that point the net force is upward, so the mass reverses.

(c) Set force, rather than speed, to zero.

$mg-kx=0$ at $$x=0.10\ \mathrm m.$$ There $$K=0.50\ \mathrm J,\qquad v=1\ \mathrm{m/s}.$$ This is the maximum speed, not a stopping point.