# Potential Energy of a Conservative Force

1. Oct 25, 2007

### Jacobpm64

1. The problem statement, all variables and given/known data
A mass $$m$$ hangs on a vertical spring of spring constant $$k$$.
(a) How far will this hanging mass have stretched the spring from its relaxed length?
(b) If you now push up on the mass and lift it until the spring reaches its relaxed length, how much work will you have done against gravity? Against the spring?

2. Relevant equations
I'll include them in my attempt at a solution.

3. The attempt at a solution
(a)
$$F = -kx$$ <--- force of a spring
$$U(x) = -W$$
$$U(x) = -\int_0^{x} -kx' dx'$$
$$U(x) = k (\frac{1}{2}x'^2)_0^{x}$$
$$U(x) = \frac{1}{2} kx^2$$ <--- potential energy of spring

The energy in the relaxed state is:
$$E_1 = \frac{1}{2}mv_1^{2} + \frac{1}{2}kx_1^{2} = \frac{1}{2}mv_1^{2} + 0$$

The energy in the stretched state is:
$$E_2 = \frac{1}{2}mv_2^{2} + \frac{1}{2}kx_2^{2} = 0 + \frac{1}{2}k(-x)^2$$

The energy conserves, so $$E_1 = E_2$$:
$$\frac{1}{2}mv_1^{2} = \frac{1}{2}kx^2$$
$$x = \sqrt{\frac{mv_1^{2}}{k}}$$ <--- How far the spring stretches

(b)
$$U(0) - U(x) = -W$$
$$-\frac{1}{2}kx^2 = -W$$
$$W_{you} = \frac{1}{2}kx^2$$
$$W_{grav} = -mg(-x) = mgx$$

The work you do against gravity is $$W_{you} - W_{grav}$$:
$$W = \frac{1}{2}kx^2 - mgx$$
$$W = x ( \frac{1}{2}kx - mg)$$ <--- work you do against gravity

The work the spring does is:
$$W_{spring} = -\frac{1}{2}kx^2$$
$$W_{you} = \frac{1}{2}kx^2$$

The work you do against the spring is $$W_{you} - W_{spring}$$ :
$$W = \frac{1}{2}kx^2 + \frac{1}{2}kx^2$$
$$W= kx^2$$ <--- work you do against the spring

I have no idea if this is correct, as the answer is not in the back of the book.. so I can't check myself on this particular problem. Please critique my work. I don't know if all of the equations make sense, as I may have been just writing stuff down and trying to force things to work. Thanks in advance.

2. Oct 25, 2007

### PhanthomJay

In part a, the mass is hanging from the spring...at rest. To calculate the spring displacement, you need only use Hooke's Law. There is no velocity associated with the system at rest.
For part b, you slowly push up the mass back to the relaxed length of the spring. The work you do, which is work done by a non conservative force, is the change in the PE of the spring (which is negative) plus the change in the PE of the mass (positive)(conservation of energy theorem where there is no change in KE).

3. Oct 25, 2007

### Jacobpm64

so my work was a bunch of rubbish haha!

Let me see if i can fix it.

so would the force be mg, so.

$$F = -kx$$
$$mg = -kx$$
$$x = \frac{-mg}{k}$$ <--- is that the spring displacement?

And part two is still not very clear to me.

4. Oct 25, 2007

### PhanthomJay

Part 'a' is correct if you consider down as the negative direction, so that's OK.
Part 2 does tend to get confusing, in some respects because you have work being done by a non conservative force (you)against 2 conservative forces (gravity and the spring forces). And the plus and minus signs will try to get the best of you. Just remember the conservation of energy theorem
$$W_{nc} = \Delta{PE_{spring}} + \Delta{PE_{gravity}} + \Delta{KE}$$
and note that there is no KE change, and $$W_{nc}$$ is the work done by you.

5. Oct 25, 2007

### Jacobpm64

and change in potential energy just equals negative work.

6. Oct 26, 2007

### PhanthomJay

No, change in PE equals negative work done by conservative forces.