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Kinetic Energy

The energy of any system is the capacity of doing any work. This is mainly classified into kinetic energy and potential energy. But it is always found according to its nature like when it is in form of heat then it’s a thermal energy, work can be done in mechanical ways etc. Thus, according to work and system, it is transform from one form to another form of energy and remains constant for system to follow the conservation law of energy. Different forms of energy are related to motion of particles.

If a body is in static state then it contains potential energy but when it is thrown then its potential energy is converted into kinetic energy. Similar to spring or in bow contains potential energy in the rest condition due to their configuration. Nuclear energy is also potential energy due to the configuration of fundamental subatomic particles that present in the nucleus of an atom. Let’s discuss the kinetic energy which is the energy of motion and its mathematical formulation with its examples.

Examples of Kinetic Energy

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What is Kinetic Energy?

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The word kinetic has a base in Greek, referring to a motion. So, in simple terms we can say that body in motion acquires some energy, which we call as kinetic energy.
Kinetic Energy definition as the Energy possessed by an object by virtue of its motion.

Kinetic Energy Units

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Dimensionally, the kinetic energy of any object is equivalent to the product of its mass and the square of its velocity. In fps system of units, the unit of kinetic energy is lbm-(ft/s)2. In SI system of units, the unit of kinetic energy becomes as kgm2/s2 better known as Joule.

We know that kinetic energy of an object i.e.
Kinetic Energy = The work done to set the object in motion from rest at a specified velocity.
Kinetic Energy = Change in the velocity of the object under motion of a different velocity.
Let us consider, an object of mass ‘m’ is at rest and given an acceleration ‘a’ to set a velocity of ‘v’ within a time ‘t’. The distance ‘s’ moved by the object in that time is the product of the average velocity and time.
Distance S = Average Velocity $\times$ time.
In other words,
S = $\frac{v – 0}{2}$ $\times$ t = $\frac{vt}{2}$.

The force F required to set the object in motion as per above condition is given by,
F = ma = $\frac{mv}{t}$

Hence, the work W done for this motion is

W = Fs = $\frac{mv}{t}$ $\times$ $\frac{vt}{2}$
= $\frac{1}{2}$ mv2

As per the definition of kinetic energy, this is the kinetic energy E acquired by the object
E = $\frac{1}{2}$ mv2
This is called as the Kinetic Energy Equation also called kinetic energy formula.

Work Energy Theorem : According to this principle, work done by a force in displacing a body, gives the measure of the change in kinetic energy of the body. i.e, The change in the kinetic energy of an object is equal to the net work done on the object.

W = K.E initial - K.E final = $\frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$
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Can kinetic Energy be Negative?

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The above question is debatable and there could be conflicting answers. But our view is that kinetic energy can not be negative. A quantity is expressed in negative terms when it is associated with a direction. Fundamentally, energy is a scalar quantity which has no direction. Further, a kinetic energy is a product of a mass and square of a velocity. There is no way that either of the term can be negative.
There are basically four types of Kinetic Energy based on its motion:
  1. Translational Kinetic energy
  2. Turbulent Kinetic energy
  3. Relativistic kinetic energy
  4. Negative kinetic energy

The Energy possessed by the body when the body is moving along the straight line is called Translational Kinetic energy.

Turbulence kinetic energy (TKE) is the mean kinetic energy per unit mass associated with eddies in turbulent flow.

Relativistic Kinetic energy is the Kinetic energy possessed by the body acquiring velocity comparable to the velocity of light.

Negative kinetic energy is the kinetic energy possessed by the body when its velocity decreases as compared to its initial Velocity.

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Average Kinetic Energy

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By default kinetic energy means only an average kinetic energy. This is because the distance for the work done is calculated on the average of final and initial velocities in a time interval.
Hence, for the translational kinetic energy (meaning the kinetic energy of a linear motion), the average kinetic energy is defined as,
E = $\frac{1}{2}$ mv2
or
average Velocity v = $\frac{V+U}{t}$
where U = initial velocity,
V = final velocity,
t = time taken.
The velocity is considered at the center of mass.

Potential and Kinetic Energy

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The most common set in transformation of energies is potential and kinetic energy. A potential energy is defined to be an energy by virtue of its position of an object. Suppose an object of mass ‘m’ is at a height of ‘h’, its potential energy P is defined as:
P = mgh,
where m is the mass,
g is the acceleration due to gravity
h is the height from ground.

Suppose the same object is dropped from that height, it reaches the ground with a velocity ’v’, thus acquiring a kinetic energy
E = $\frac{1}{2}$ mv2

At this point the entire potential energy is dissipated and as per conservation of energy, P.E = K.E

mgh = $\frac{1}{2}$ mv2

or

h = $\frac{v^{2}}{2g}$
Which gives an important relation between the height and the final velocity of falling objects.

Relativistic kinetic Energy

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In the above discussion of kinetic energy, we assumed velocity of an object that can be negligible when compared to the velocity of light. The study with such assumption is called classical mechanics.

But after the introduction of theory of relativity, a review becomes necessary when the velocity of objects in motion is comparable with velocity of light. In such a case the mechanics is called Relativistic Mechanics.

In relativistic mechanics, the kinetic energy is upgraded as Relativistic Kinetic Energy.
The velocity of the object considered for determining the relativistic kinetic energy is a function of the ratio of the actual velocity to the velocity of light. The relativistic kinetic energy of a rigid object of mass m with actual velocity v is defined as,
E = mc2(k-1)
where,
c is the velocity of light and
k is given by the relation k2 = $\frac{1}{(1 – (\frac{v}{c})^{2})}$

Examples of Kinetic Energy

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We see a number of practical examples of kinetic energies in real life:
  1. When you hit a nail on a wall with a hammer with a force, thereby, with a velocity, the kinetic energy of the hammer helps you to drive the nail into the wall.
  2. You are able to travel in a car, train or an airplane because of the kinetic energy of the vehicle. Your body is made to travel a distance because you take a part of the kinetic energy of the vehicle.
  3. The kinetic energy of the gushing water from a dam transforms into rotational energy of an electric generator which in turn generates electrical power.

Kinetic Energy Problems

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Following are some Kinetic Energy Problems:

Solved Examples

Question 1: A truck and a car are moving with the same kinetic energy on a road. Their engines are simultaneously switched off. Which one will stop at a lesser distance?
Solution:
 
The vehicle stops when its kinetic energy is spent in working against the force of friction between the tyres and the road. This force of friction varies directly with weight of the vehicle.
As the K.E. = work done = Force of friction $\times$ distance
              E = F $\times$ d
              or
              d = $\frac{E}{F}$.
where E = Kinetic energy,
          F = Force
For given kinetic energy, distance s will be smaller, where F is larger, such as in case of truck. Thus truck stops Earlier.   

 

Question 2: A body of mass 75 kg has a momentum of 1500kg ms-1. Calculate its kinetic energy?

Solution:
 
As momentum p = m $\times$ v.
where m = Mass = 75 kg,
          v = Velocity
          p = 1500 kg ms-1
     1500 kg ms-1 = 75 kg $\times$ v
                       v = 20m/s.
Kinetic energy, K.E. = $\frac{1}{2}$ mv2
                              = $\frac{1}{2}$ $\times$ 75kg $\times$ 400 m2/s2
                              = 15000 Joules.
 

Question 3: Given that the displacement of body in meters is a function of time as follows: x = 2t4 + 5. The mass of the body is 2kg. What is the increase in its kinetic energy one second after the start of the motion?
Solution:
 
Velocity = v = $\frac{dx}{dt}$ = $\frac{d}{dt}$ (2t4 + 5) = 8t3
After one second velocity is 8(1)3 = 8 m/s
Kinetic energy, K = $\frac{1}{2}$ mv2
where m = Mass of the body,
          v = Velocity
K  = $\frac{1}{2}$ $\times$ 2 kg $\times$ (8)2 m2/s2
    = 64J.
 

Question 4: If we throw a body upwards with velocity of 4ms-1, then at what height its kinetic energy reduces to half of the initial value? Take g = 10 ms-2.
Solution:
 
Initial energy = $\frac{1}{2}$ m(4)2 = 8m m/s2.
Let kinetic energy at a height h be $\frac{8}{2}$ = 4m m/s2.
It will also be equal to the potential energy.
Hence K.E. = P.E
$\frac{1}{2}$
mv2 = mgh.
4m m/s2 = m $\times$ 10 m/s2 $\times$ h
h = 0.4m.
 

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