Potential energy is the energy that an object has stored in it due to its position. When we think of potential energy, often the first thing that we can imagine is an object high in the air and just starting to fall. It is having potential energy stored in it due to its height. This energy will be turned into kinetic energy as it falls. However, there are some other situations in which an object can have potential energy. One such example is an elastic material. In this article, we will discuss the elastic potential energy formula with an example. Let us learn the concept!
Elastic Potential Energy Formula
What is elastic potential energy?
This is the energy than an object has in it due to being deformation of its shape. Any object which can be deformed and then return to its original shape, then it can have elastic potential energy. Examples of such objects are rubber bands, sponges, and bungee cords, and many others.
When we deform these objects, they move back to their original shape on their own. It is only possible due to accumulated potential energy which is elastic potential energy. Thus, elastic potential energy is the stored energy of a compressible or stretchable object.
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Elastic Potential Energy Formula
We can compute Elastic potential energy by using fundamental formula as below:
Elastic potential energy = force $$\times$$ displacement.
It is computed as the work done to stretch the spring which depends on the spring constant k and the displacement stretched.
According to Hooke’s law, the force applied to stretch the spring is directly proportional to the amount of stretch. In other words,
The force required to stretch the spring is directly proportional to its displacement. It is given as
P.E. = Magnitude of Force × Displacement
P.E. = \( \frac{1}{2}kx^2 \)
-ve sign indicates the opposite direction.
Where,
P.E. | Elastic Potential Energy |
k | Spring Constant |
x | Displacement stretched |
dx | Small displacement |
This gives the elastic Potential Energy expressed in Joule.
Solved Examples
Q.1: A compressed spring has a potential energy of 50 Joule and its spring constant is 200 N/m. Calculate the displacement of the spring due to this potential energy.
Solution:
Given parameters in the problem:
Potential energy P.E = 50 J,
Spring Constant, k = 200 N/m,
The Potential energy formula is given by:
P.E. =Â \( \frac{1}{2}kx^2 \)
Rearranging the equation, we get
x = \( \sqrt( \frac {2× P.E.}{ k } ) \)
Now, substituting the values,
x =\( \sqrt( \frac {2\times 50 }{ 200 } ) \)
x=Â \( \sqrt(\frac {1}{2} ) \)
x =Â \( \frac{1}{\sqrt 2} meter \)
x = 0.707 meter
Therefore displacement will be 0.707 meter.
Q.2: The vertical spring is linked with a load of mass 5 kg which is compressed by 10m. Find out the Force constant of the spring.
Solution:
Given: Mass m = 5 kg
Distance x = 10 cm
Force formula is given by
F = ma
= 5 kg × 9.8 m/s²
F = 49 N
Also, Force in the stretched spring is
F = k x
Force Constant k is given by
k = \( \frac{F}{x}\)
= \( \frac{49}{10}\)
k = 4.9 N/m
Therefore spring constant is 4.9 N/m.
Typo Error>
Speed of Light, C = 299,792,458 m/s in vacuum
So U s/b C = 3 x 10^8 m/s
Not that C = 3 x 108 m/s
to imply C = 324 m/s
A bullet is faster than 324m/s
I have realy intrested to to this topic
m=f/a correct this
Interesting studies
It is already correct f= ma by second newton formula…