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## Class 11 Physics (India) - Hindi

### Course: Class 11 Physics (India) - Hindi > Unit 4

Lesson 5: Spring potential energy and Hooke's law (Hindi)- Intro to springs and Hooke's law (Hindi)
- Potential energy stored in a spring (Hindi)
- Spring potential energy example (Hindi)
- Calculating spring force
- Calculating elastic potential energy
- Spring potential energy and Hooke's law review
- What is Hooke's Law?
- What is elastic potential energy?

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# What is elastic potential energy?

Learn what elastic potential energy means and how to calculate it.

# What is elastic potential energy?

*Elastic potential energy*is energy stored as a result of applying a force to deform an elastic object. The energy is stored until the force is removed and the object springs back to its original shape, doing work in the process. The deformation could involve compressing, stretching or twisting the object. Many objects are designed specifically to store elastic potential energy, for example:

- The coil spring of a wind-up clock
- An archer's stretched bow
- A bent diving board, just before a divers jump
- The twisted rubber band which powers a toy airplane
- A bouncy ball, compressed at the moment it bounces off a brick wall.

An object designed to store elastic potential energy will typically have a high elastic limit, however all elastic objects have a limit to the load they can sustain. When deformed beyond the elastic limit, the object will no longer return to its original shape. In earlier generations, wind-up mechanical watches powered by coil springs were popular accessories. Nowadays, we don't tend to use wind-up smartphones because no materials exist with high enough elastic limit to store elastic potential energy with high enough energy density.

# How can we calculate elastic potential energy for an ideal spring?

Our article on Hooke's law and elasticity discusses how the magnitude of the force $F$ due to an ideal spring depends linearly on the length it has been compressed or expanded $\mathrm{\Delta}x$ ,

where $k$ is some positive number known as the spring constant. The spring force is a conservative force and conservative forces have potential energies associated with them.

From the definition of work we know that the area under a force vs displacement graph gives the work done by the force. Figure 1 shows a plot of force vs displacement for a spring. Because the area under the curve is a triangle and no energy is lost in an ideal spring, the elastic potential energy $U$ can be found from the work done

**Exercise 1:**A truck spring has a spring constant of

**Exercise 2a:**A trained archer has the ability to draw a longbow with a force of up to 300 N, extending the string back by 0.6 m. Assuming the bow behaves like an ideal spring, what spring constant would allow the archer to make use of his full strength?

**Exercise 2b:**What potential energy is stored in the bow when it is drawn?

**Exercise 2c:**Assuming the arrow has a mass of 30 g, approximately what speed will it be fired at?

**Exercise 2d:**Suppose that measurements from a high speed camera show the arrow to be moving at a somewhat slower speed than predicted by conservation of energy. Is there any work being done that we have not accounted for?

# What about real elastic materials?

In our article on Hooke's law and elasticity we discuss how real springs only obey Hooke's law over some particular range of applied force. Some elastic materials such as rubber bands and flexible plastics can function as springs but often have hysteresis; this means the force vs extension curve follows a different path when the material is being deformed compared to when it is relaxing back to its equilibrium position.

Fortunately, the basic technique of applying the definition of work that we employed for an ideal spring also works for elastic materials in general. The elastic potential energy can always be found from the area under the force vs extension curve, regardless of the shape of the curve.

In our earlier analysis, we have considered the ideal spring as a one-dimensional object. In reality, elastic materials are three dimensional. It turns out that the same procedure still applies. The equivalent to the force vs extension curve is the

*stress vs strain*curve.Where a three-dimensional elastic material obeys Hooke's law,

**Exercise 3:**Figure 3 shows a stress vs strain plot for a rubber band. As it is stretched (loaded), the curve takes the upper path. Because the rubber band is not ideal, it delivers less force for a given extension when relaxing back (unloaded). The purple shaded area represents the elastic potential energy at maximum extension. The difference in area between the loaded and unloaded case is shown in yellow. This represents the energy which is lost to heat as the band is cycled between stretched and relaxed.

If the rubber band has length $100\text{}\mathrm{mm}$ , width $10\text{}\mathrm{mm}$ and thickness $1\text{}\mathrm{mm}$

**how much heat is generated in the band as it is stretched and released?**# Attributions

## Want to join the conversation?

- In the rubber band example, is the heat dissipated as work is done stretching the rubber band, or as the rubber band is being unloaded?(13 votes)
- I'm fairly new to this topic, but from past experience of doing this in 3rd grade, we used to stretch a rubber band really quickly, then put it to our upper lip (while it was still stretched.). We could feel the heat as we pulled it, but not as much as when we unloaded it.

So mathematically, I can't tell you the answer, but from experimentation, it does produce heat when loaded.(33 votes)

- Exercise 2 is worded very strangely. It sounds like 0.6m is just the distance the string gets pulled back when 300N is applied, which would imply a specific spring constant, so why does the question make it sound like the spring constant could be anything? Is 0.6m just the maximum limit to how far the bow can be pulled back? That should be stated more clearly. Also, wouldn't any spring constant greater than 500N/m also allow the archer to use his full strength?(7 votes)
- The way I understood it, 300N is his maximum strength. If he useed 250N and produced an extension of 0.6m, the spring constant would be different (in which case the bow would probably be made in a different shape or size or with a different material).(2 votes)

- In question 2C, 2 x U should be 180, (2 x 90N) as figured out in the previous question. So how does 2 x U = 2.9? Someone please explain, thanks.(6 votes)
- Dude it not 2.9. Its 2*90. The dot there is for multiplication(3 votes)

- Why in Exercise1 250J/spring = 1000J? ( solution)(3 votes)
- There are four springs on the truck in exercise 1 (one per wheel.) The displacement given is the displacement of the entire truck, meaning each individual spring is compressed 0.1 m. The calculation done (PE=(0.5)(5*10^4)(0.1)^2) gives you the amount of energy stored in each individual spring. Therefor the total energy stored in all four springs is 250 J * 4 springs = 1000 J total.(7 votes)

- In question 3, why is the heat energy = stress * strain * volume, instead of stress* strain * volume * .5, or am I missing something?(6 votes)
- In a stress-strain graph, is the stress plotted always (force applied) / (original cross-sectional area of material) or is it (force applied) / (cross-sectional area of material when that force is applied)? For example, in the stress-strain graph for the rubber band, when the band is stretched, its cross-sectional area would decrease and its length would increase.(4 votes)
- prove how energy/volume =1/2 stress.strain(4 votes)
- In the extension vs force graph, what if the force was always constant? (e.g. the weight of a ball pulling down a vertical spring). In the graph, it isn't and just keeps growing as the displacement grows.(1 vote)
- If the force was constant, you wouldn't have a spring. That's not what springs do.(5 votes)

- i don't understand how exercise 3 went from 0.05N/mm^2 to 5 x 10^4 N/m^2.(2 votes)
- Why do we multiply the volume of the rubber by the heat in the last exercise?

Thanks(2 votes)- Just above exercise 3 it states that

"Where a three-dimensional elastic material obeys Hooke's law,

Energy/volume=(Stress*Strain)/2"

The rubber band we are using for this exercise is a 3-dimensional object and figure 3 is a stress vs strain plot. Notice that the first part of the equation is Energy/volume. When you find the area of the yellow part that corresponds to heat you get an expression of (Energy that is spent on heat)/(Volume of the object). So in order to find the energy spent on heat we have to multiply by the volume.(2 votes)