# Strange factor of 2 in Hooke's Law [duplicate]

I was working on a PGRE practice problem:

A brick of mass $$m$$ falls onto a masses spring with spring constant $$k$$ from a height $$h$$ above it. What is the maximum distance the spring will be compressed from its equilibrium length?

I use conservation of energy and get that $$x = \frac{mg}{k}(1+\sqrt{1+\frac{2kh}{mg}})$$ with the book confirms is correct. However, in the limiting case that h goes to zero, this reduces to $$x=\frac{2mg}{k}$$ (which the book also notes is correct). If h is zero though, the problem is just find the equilibrium of a spring of spring contact k supporting a brick of mass m, in which case doesn't Newton's second law just give $$mg - kx = 0$$ implying $$x = \frac{mg}{k}$$?

I also tried to solve this by saying that when the brick is initially on the spring (in the case that h=0) then the system has gravitational potential energy of zero. Once the spring reaches its equilibrium, the energy will be $$0.5kx^2 - mgx=0$$ which also gives me $$x=\frac{2mg}{k}$$

I seem to recall something in my first year physics course about the average force of the spring, but I can't remember anything definitive. Can anyone explain where this factor of two comes from?

• Sorry, should have looked for duplicates first – Aaron Stevens Oct 8 '19 at 5:35
• – Farcher Oct 8 '19 at 7:05

No it is not. The place where the mass comes to rest will not be the equilibrium position. At equilibrium the mass will still have kinetic energy, so it will overshoot equilibrium and move farther to the $$2mg/k$$ that you have found to be the correct answer. Then the mass will move upwards and oscillations will continue indefinitely.
Remember, equilibrium is where forces are equal to $$0$$, not where the velocity is equal to $$0$$. This also means you need to change your wording for your final method of solution. i.e. instead of saying "Once the spring reaches its equilibrium..." You should say "Once the mass is at rest..."