Kinetic Energy for Class 9 th

Life is impossible without energy. The demand for energy is ever increasing. Where do we get energy from? The Sun is the biggest natural source of energy to us. Many of our energy sources are derived from the Sun. We can also get energy from the nuclei of atoms, the interior of the earth, and the tides.

            The word energy is very often used in our daily life, but in science we give it a definite and precise meaning.

            The object which does the work loses energy and the object on which the work is done gains energy.   

            Any object that possesses energy can do work.

            The energy possessed by an object is thus measured in terms of its capacity of doing work. The unit of energy is, therefore, the same as that of work, that is, joule (J). 1 J is the energy required to do 1 joule of work. Its unit is Kgm²s^-2

            Joule, unit of work or energy in the International System of Units (SI) is equal to the work done by a force of one newton acting through one metre. We may also say,

            It is equal to the energy transferred to an object when a force of one newton acts on that object in the direction of the force’s motion through a distance of one metre.

            Sometimes a larger unit of energy called kilo joule (kJ) is used. 1 kJ equals 1000 J.

Forms of Energy

            There are various forms of energy few of which are Mechanical energy (which may be Potential energy, or Kinetic energy), Heat energy, Chemical energy, Electrical energy and Light energy.

            As per level of class 9th we will discuss the very important 2 types among them. Kinetic & Potential. In this session we take up Kinetic Energy.

Kinetic Energy

            A moving object can do work. An object moving faster can do more work than an identical object moving relatively slow. A moving bullet, blowing wind, a rotating wheel, a speeding stone can do work.

            How does a bullet pierce the target?  Energy gained by it due to its speed makes its way.

            How does the wind move the blades of a windmill?  The blades interfering in the way of the wind bear its pressure. When the blades obstruct the path of wind, energy of wind gets transferred to it, in turn moving them.

             Objects in motion possess energy. We call this energy kinetic energy.  Kinetic energy is the energy possessed by an object due to its motion. The kinetic energy of an object increases with its speed.  We may say that the kinetic energy of a body moving with a certain velocity is equal to the work done on it to make it acquire that velocity.

            A falling coconut, a speeding car, a rolling stone, a flying aircraft, flowing water, blowing wind, a running athlete etc. possess kinetic energy.

Remember it by Heart that Kinetic Energy is possessed by a moving object only. As soon as it comes to rest Kinetic Energy becomes zero.

Derivation of Kinetic Energy Formula

            Consider an object of mass, m moving with a uniform velocity, u. Let it now be displaced through a distance s when a constant force, F acts on it in the direction of its displacement.

            Let us recollect that Work done, W = F s.

            The work done on the object will cause a change in its velocity. Let its velocity change from u to v.  Where v is its final velocity.  

            When velocity has changed there would be either acceleration or a retardation depending upon the Force whether it is positive or negative.

            Let the acceleration produced be a.

            Let us recollect the three equations of motion studied in lesson 8:

v = u + at

s = ut + ½ at²

2 as = v² – u²

            Out of them can you point out the relevant equation which uses all the required variables, a, s, u and v.

            Yes you are correct.  It is the third one.  Now choosing it out:

2 as = v² – u²

Gives us,

s = (v² – u² )/ 2a

            In the lesson upon Work we have seen that W = Fs  & F = ma

So, we can write,

W = mas (read as m into a into s)

Replacing s in this equation by the value of s derived above, we may write,

W = ma (v² – u²) / 2a

  Or,

W = ½ m (v² – u²)

For ease considering the initial velocity u to be zero (that is when the object was stationary at the beginning), we get the equation,

W = ½ m v²

                    As the object gets energised only by the work which is done upon it, we may conclude that the Energy acquired by the object can be represented by the same equation as above i.e. half mv square.  Here in this case the only form of energy the object possesses is Kinetic Energy.

So,

Kinetic Energy, K.E. or E_k  = ½ m v²

Exa-  An object of mass 15 kg is moving with a uniform velocity of 4 ms^–1. What is the kinetic energy possessed by the object?

Solution: Mass of the object, m = 15 kg, velocity of the object, v = 4 ms^–1

Placing the values in the equation, E_k  = ½ m v²  we get,

E_k  = ½  × 15 kg × 4 ms^–1 × 4 ms^–1 = 120 J.  Hence the kinetic energy of the object is 120 J.

Exa-  What is the work to be done to increase the velocity of a car from 30 km h^–1 to 60 km h^–1 if the mass of the car is 1500 kg?

Solution: 

            Mass of the car, m =1500 kg,

        Initial velocity of car, u = 30 km h^–1 = 30 ×1000m / (3600 s) = 8.33 m s^–1.

[Why did we do it?

Because we have to change the units into their S.I. units before we proceed to calculate. And SI unit of velocity is ms^-1 and notkm h^–1  .  So km is multiplied by 1000 to change it into meter and hour by 3600 to convert it into seconds. You know that 1 hour has 60 minutes and a minute has 60 seconds]

Similarly,

Final Velocity, v = 60 km h^–1  = 60 ×1000m / 3600 s = 16.66 m s^–1.

So,

Initial Kinetic Energy, E_ki  = ½ m u²  = ½ x 1500 kg x (8.33 m s^–1)² 

= 52,041.675 J

            Final Kinetic Energy, E_kf  = ½ m v²  = ½ x 1500 kg x (16.66 m s^–1)² 

= 208,166.7 J

So,

            The Work done = Change in KE = E_kf – E_ki

                           = 208166.7 – 52041.675 = 156125.025 J  = 156 KJ (App.)

Question 3 of Page 152

            The kinetic energy of an object of mass, m moving with a velocity of 5 m s^–1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?

Solution:

Let the mass be m; v = 5 ms^-1  &  K.E = 25 J
as we know,

            K.E = ½ m v²
25 = ½ m (5)²
25 x 2 = 25m
Therefore, m = 2 kg

Now if the velocity becomes 2 times then the velocity is 2 x 5 = 10 ms^-1
So the new K.E = ½  x 2 x 10²
= 100 J

Now if the velocity becomes 3 times then the velocity is 3 x 5 = 15 ms^-1
So the new K.E = ½  x 2 x 15²
= 225 J

[Did you notice that by doubling the velocity 2 folds the energy increases from 25 Joules to 100 Joules i.e. 4 times increase; & by increasing the velocity 3 times the energy increases to 225 Joules.  It is an increase of 9 fold. Can’t we relate it to the unprecedented results by people who increase their efforts.  Can’t we exert ourselves a bit more to achieve much higher goals!!]