Elastic Potential Energy Formula and Free Fall Formula

Elastic Potential Energy Formula

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

×

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. = 12kx2

-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. =  12kx2

Rearranging the equation, we get

x =  (√2×P.E.k)

Now, substituting the values,

x =(√2×50200)

x=  (√12)

x =  12√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 = Fx

= 4910

k = 4.9 N/m

Therefore spring constant is 4.9 N/m.

Free Fall Formula

Freefall is a common kind of motion which everybody can observe in daily life. If we drop something accidentally we can see its motion. In the beginning, it will have low speed and until the end, it gains speed and before the collision, it reaches its maximum speed. Many factors are there to affect the speed of the object while it is in free fall. We deal with such free-fall motion and free fall formula with examples in this article. Let us learn the concept!

Free Fall Formula

Concept 

Freefall refers to a situation in physics where the only force acting on an object is gravity and hence acceleration due to gravity. Freefall as its term says is a body falling freely because of the gravitational pull of the earth. This motion will have the effect of acceleration due to gravity. This type of motion will follow the three equations of motion under gravity.

Projectile motion is another important category of free-fall problems. Although these events unfold in the three-dimensional world, for basic physics purposes, they are considered as two-dimensional on paper.

A very unique but interesting property of the acceleration due to gravity is that it is the same for all masses. This was far from the self-evident fact, until the days of Galileo Galilei. That was because in reality gravity is not the only force acting as an object falls, and the effects of air resistance tend to cause lighter objects to accelerate more slowly. It is something that we have all noticed when comparing the fall rate of a rock and a feather.

Galileo conducted this ingenious experiments at the “leaning” Tower of Pisa and proving by dropping masses of different weights from the top of the tower that gravitational acceleration is independent of the mass of the objects.

Free-fall physics problems are having the assumption of the absence of air resistance. But, in the real world, the Earth’s atmosphere provides some resistance to an object in free fall. Also, particles in the air collide with the falling object, which results in transforming some of its kinetic energy into thermal energy. This results in “less motion” or a more slowly increasing downward velocity.

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The formula for free fall:

Imagine an object body is falling freely for time t seconds, with final velocity v, from a height h, due to gravity g. It will follow the following equations of motion as:

1. h= 12gt2
2. v²= 2gh
3. v=gt

Where,

h	Height traveled
v	Final velocity
g	Acceleration due to gravity
t	Time taken

These equations can be derived from the usual equations of motions as given below, by substituting

initial velocity u=0,

distance traveled s=h and

acceleration, a=g.

We can see it as follows:

s= ut+12at2

v² =u²+ 2as

v=u+at

Freefall is the autonomous phenomena of the body with some mass. It only depends on height from the surface and the time period for which the body is flung.

Solved examples:

Example-1: Compute the height of the body if it has a mass of 2 Kg and touches the ground after 5 seconds?

Solution:

Given parameters are:

Time t = 5 sec

We have to compute the height. So, we can apply the first equation as given above.

i.e. h=  12gt2

Substituting the values,

h= 12gt2

h=  12×9.8×52

h= 4.9 × 25

h = 122.5 m

Therefore height as required will be 122.5 meter.