# Calculating Strain Energy & Potential Energy Loss: 8kg Mass on 4.2cm Wire

• Iamtoast
In summary, the mass lost 3.3 J of potential energy as it was lowered by the person placing it on the wire.
Iamtoast
A simple problem: a mass of 8 kg is placed on the end of a copper wire. The wire extends a distance of 4.2 cm. Calculate the potential energy stored in the wire and the loss of potential energy of the mass.

The elastic strain energy stored in the wire = 1/2 . F . x = 1.65 J
The loss of gravitational potential energy by the mass = mgh = 3.3 J

So where has the other 1.65 J gone? In stretching, the entire wire (with the exception, presumably, of an infinitely thin cross section immediately adjoining the support) descends by varying amounts and itself loses g.p.e.. How is energy conserved? I can only conclude that the wire in descending, gains kinetic energy which is then dissipated as thermal energy. Comments please on this please especially on mechanisms involved.

Perhaps you should check your expression for elastic potential energy.

I naturally checked this before posting. Do you have an alternative suggestion?

Iamtoast said:
How is energy conserved? I can only conclude that the wire in descending, gains kinetic energy which is then dissipated as thermal energy. Comments please on this please especially on mechanisms involved.
Yes. Think about what happens when you drop an object suspended by a spring. It bounces up and down for some amount of time. How long depends on things like air resistance and the properties of the spring. Either way, after a while, these forces damp the oscillation until it stops. Whatever is lost eventually ends up as heat.

I am not sure that this answers my question. Imagine doing this experiment in a vacuum, carefully loading the wire so that it does not oscillate. The problem here seems to be that irrespective of the size of the mass, only half of the gpe lost is recoverable as stored elastic potential energy.

Assuming that the mass was gently placed placed at the end of the wire, then it would be the hand of the person placing it there that would have taken up the missing energy. That's different than attaching the mass and letting it drop--in which case the mass would oscillate until internal losses made it stop and the missing energy would end up as thermal energy (as russ explained).

Whenever you gently hang something from a spring, the additional elastic PE is only half as much as the drop in gravitational PE of the mass.

Doc Al said:
Assuming that the mass was gently placed placed at the end of the wire, then it would be the hand of the person placing it there that would have taken up the missing energy.

How? Why?

Doc Al said:
Whenever you gently hang something from a spring, the additional elastic PE is only half as much as the drop in gravitational PE of the mass.

Again, why?

Let's work through an example with a mass on a spring. For me to gently place the mass on the spring, I have to make sure that at any point, from x=0 to x=mg/k, the net force on the mass is essentially zero--I don't want to drop it, which will make the mass oscillate. That means I have to exert an additional upward force ranging from mg (at x = 0) to 0 (at x = mg/k) to make up the difference between the mass's weight (mg) and the spring force (kx). That upward force does negative work on the spring/mass system as the mass is lowered, sucking away some of its energy. (Alternatively, you can think of the mass pushing down on my hand doing positive work on my hand as I lower it.)

You can figure out the work done by the hand directly (it's the mirror image of the work done by the spring force, so the energy equals the spring energy) or you can just figure out the net result:
Maximum stretch of spring: x = mg/k
Energy stored in spring: 1/2kx^2 = 1/2(mg)^2/k
Drop in gravitational PE: mgx = (mg)^2/k

So: Half of the gravitational energy went to the spring; half went into whatever system gently lowered the mass. Make sense?

I'm not sure I see what you are missing. A spring does not provide a constant force because it is a spring. So when you start to lower it, FxD gives you nothing because the spring is providing no force, while your hand is holding up all of the weight and providing (or absorbing, depending on your frame of reference) all of the work. The more you lower it, the less force your hand provides and the more the spring provides.

Thank for your illuminating response.

Iamtoast said:
A simple problem: a mass of 8 kg is placed on the end of a copper wire. The wire extends a distance of 4.2 cm. Calculate the potential energy stored in the wire and the loss of potential energy of the mass.

The elastic strain energy stored in the wire = 1/2 . F . x = 1.65 J
The loss of gravitational potential energy by the mass = mgh = 3.3 J

So where has the other 1.65 J gone?
HINTS:

1. If there were no spring at all, and you just dropped a mass, it's PE would decrease from its initial value. Where does all this PE go?

2. At what points does an oscillator have zero speed, the turning points or the equilibrium point?

## 1. What is strain energy?

Strain energy is the energy stored in an object when it is deformed or stretched from its original shape.

## 2. How is strain energy calculated?

Strain energy can be calculated by multiplying the force applied to an object by the distance it was deformed or stretched.

## 3. What is potential energy loss?

Potential energy loss is the decrease in the amount of energy an object has due to external forces acting upon it.

## 4. How is potential energy loss determined?

Potential energy loss can be determined by calculating the difference in potential energy before and after a change in the system.

## 5. Can strain energy and potential energy loss be calculated for any object?

Yes, strain energy and potential energy loss can be calculated for any object as long as the necessary parameters such as mass, distance, and external forces are known.

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