# Homework Help: Odd result in simple mechanics problem

1. Dec 16, 2012

### gsmith

1. The problem statement, all variables and given/known data

A weird conundrum I came across while doing some basic mechanics, I just can't figure out.

Okay, so a mass is hanging vertically from a spring and the spring is stretched a distance x.

2. Relevant equations

1/2kx2=mgh...(h=x) therefore...1/2kx2=mgx

mg=kx

3. The attempt at a solution

solve the top equation for m and you get

m=(kx)/(2g)

solve the bottom equation for m and you get

m=(kx)/g

why are they different, what am I missing or not doing correctly?

2. Dec 16, 2012

### SammyS

Staff Emeritus
The right-hand side of this equation is the gravitational Potential Energy of the mass when it is at the location where the spring is unstretched. You could say the right-hand side is the total PE, due to both gravity and the spring at the location where the spring is unstretched.

The left-hand side of this equation is the PE of this system when the spring is stretched a distance x.

What makes you think these should be equal?

3. Dec 16, 2012

### gsmith

If I define h=0 as the lowest point that the spring stretches to, then the change in potential energy would equal mgh where h equals the distance that the spring has stretched. This gravitational potential energy lost is transferred to the spring, whose energy is describes by the equation 1/2kx^2. Due to the conservation of mechanical energy these two values must be equal.

4. Dec 16, 2012

### TSny

Suppose you release the mass at the unstretched position of the spring (x = 0), then it will fall until the spring stretches to a point x2 where the mass will be instantaneously at rest. But x2 is not the equilibrium position x1 where the mass would hang at rest. At x2 it would be true that mgx2 = kx22/2. But at x2 the mass has passed through the equilibrium position (x1) and the spring force at x2 is greater than the gravitational force. So the mass starts moving back upwards toward the equilibrium position (it's oscillating in SHM).

If you start the mass at the unstretched position of the spring (x = 0) and you slowly lower the mass to the equilibrium position (x1), then you do work on the system with the force that you apply to the mass. So, you cannot assume that the loss in gravitational PE equals the gain in spring PE.

Last edited: Dec 16, 2012