Calculating Spring Potential Energy in a Total Energy System

[flyingpig]

Homework Statement

A 0.25 kg object is suspended on a light spring of spring constant 49 N/m. The spring is then compressed to a position 15 cm above the stretched equilibrium position. How much more energy does the system have at the compressed position than at the stretched equilibrium position?

The Attempt at a Solution

[PLAIN]http://img35.imageshack.us/img35/3553/unledbg.jpg

$$E_{total} = PE_{spring} + PE_{regular potential energy}$$

So before it has been stretched by the mass

$$mg = ky$$

$$y = mg/k = (0.25kg)(10m/s^2)/(49N/m) = 0.051m$$

Now I am stuck...

Is the total energy

$$\sum E_i = mgy$$

$$\sum E_i = \frac{1}{2}ky^2$$

$$\sum E_i = \frac{1}{2}ky^2 + mgy$$

 

[PeterO]

Homework Statement

A 0.25 kg object is suspended on a light spring of spring constant 49 N/m. The spring is then compressed to a position 15 cm above the stretched equilibrium position. How much more energy does the system have at the compressed position than at the stretched equilibrium position?

The Attempt at a Solution

[PLAIN]http://img35.imageshack.us/img35/3553/unledbg.jpg

$$E_{total} = PE_{spring} + PE_{regular potential energy}$$

So before it has been stretched by the mass

$$mg = ky$$

$$y = mg/k = (0.25kg)(10m/s^2)/(49N/m) = 0.051m$$

Now I am stuck...

Is the total energy

$$\sum E_i = mgy$$

$$\sum E_i = \frac{1}{2}ky^2$$

$$\sum E_i = \frac{1}{2}ky^2 + mgy$$

#1 Note that the only reason I would give a spring constant of 49 was if I expected people to use a g value of 9.8

In the first equilibrium position, there will be a certain amount of elastic energy in the spring plus a gravitational potential energy.

If you took your reference height to be that equilibrium position then your second alternative would be the total energy.

When you lift the mass 15cm, you increase the gravitational energy [mgh] and reduce the elastic potential energy because the spring is being stretched 15 cm less.

Note: the description :
"The spring is then compressed to a position 15 cm above the stretched equilibrium position." is a little misleading I feel, as the spring will still be stretched compared to an unladen spring, just not stretched as much as it was when the 0.25 kg was hung from the spring.

 

[modulus]

Note that the initial position of the system is the equilibrium position. That means, for the system to move up by 15cm, an external force will have to act on the system, and thus, the total energy of the system (block-spring-earth system) will necessarily change.

Now, look carefully at what the question asks you:
"How much more energy does the system have at the compressed position than at the stretched equilibrium position"
It is asking you what that change in energy is...not the final energy of the system. Also note that the block will not have any KE after the transition (though not explicitly stated in the question, that is what it seems to imply...otherwise the system of equations formed could not be solved).

 

[flyingpig]

Sorry for my late reply guys

When you lift the mass 15cm, you increase the gravitational energy [mgh] and reduce the elastic potential energy because the spring is being stretched 15 cm less.

But when you compress it back, don't increase the stored elastic potential energy?

Note that the initial position of the system is the equilibrium position. That means, for the system to move up by 15cm, an external force will have to act on the system, and thus, the total energy of the system (block-spring-earth system) will necessarily change.

I am confused, what do you mean by the equilibrium position? Do you mean the spring's equilibrium position?

I don't think it was at the spring's equilibrium point.

So I still have

mg = ky

mg/k = y

$$E_i = \frac{1}{2}ky^2 + mgh$$

$$E_f = \frac{1}{2}ky'^2 + mgh_f$$

$$\Delta E = mg\Delta h + \frac{k}{2}(y'^2 - y^2) = 0.25*9.8*(15cm) + \frac{k}{2}{15^2 - (\frac{mg}{k})^2}$$

 

[cepheid]

Hi flyingpig,

I am confused, what do you mean by the equilibrium position? Do you mean the spring's equilibrium position?

I don't think it was at the spring's equilibrium point.

The equilibrium position is the position at which the forces on the mass are balanced (zero net force). In the case of a horizontal spring, since no forces are present other than the spring's restoring force, equilibrium occurs when the spring is neither stretched nor compressed, and hence the restoring force is zero.

In the case of a vertical spring, you also have gravity. So, it's no longer true that equilibrium occurs when the spring is at its normal length. Instead, equilibrium occurs when the spring is stretched sufficiently that the upward restoring force balances the weight of the suspended mass. Your problem calls this equilibrium point the stretched equilibrium position to emphasize that equilibrium occurs when the spring is stretched (unlike in the horizontal case). It is this stretched equilibrium point that PeterO was referring to as the initial position.

So I still have

mg = ky

mg/k = y

$$E_i = \frac{1}{2}ky^2 + mgh$$

$$E_f = \frac{1}{2}ky'^2 + mgh_f$$

$$\Delta E = mg\Delta h + \frac{k}{2}(y'^2 - y^2) = 0.25*9.8*(15cm) + \frac{k}{2}{15^2 - (\frac{mg}{k})^2}$$

Everything looks fine except the difference in spring extensions. Initially, the spring is extended to a length of y = 5.1 cm beyond the unstretched length. (you worked this out). But then it gets pushed upward by 15 cm, which means the new length must be y' = 5.1 - 15 = -9.9. So the spring becomes 9.9 cm shorter than its unstretched length. These displacements y and y' are what you enter into the elastic potential energy terms.

EDIT: the 5.1 becomes exactly 5.0 if you use g = 9.8, as PeterO pointed out.