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What is kinetic energy? Prove that the kinetic energy of a body is \(\frac{1}{2} m v^{2}\) Derive the relation for work-energy theorem while explaining it.

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Kinetic Energy
The energy associated with the motion of an object is called kinetic energy. For example, the motion of a falling stone, motion of a car, etc. The kinetic energy of an object is a measure of the work that an object can do by the virtue of its motion.

Consider a body having mass m initially at rest (u = 0);initial velocity u = 0. Let a constant force Facts on a body due to which it gains a velocity v after travelling a displacement s. Then, kinetic energy of the body;
K = W = Fs ………………………….(1)
By Newton’s law of motion
F = ma ………………….(2)
From the third equation of motion
\(\begin{array}{l}{v^{2}=u^{2}+2 a s} \\ {v^{2}=(0)^{2}+2 a s} \\ {s=\frac{v^{2}}{2 a}}\end{array}\)
from the equation (1), (2) and (3)
\(\therefore \qquad K=W=\max \frac{v^{2}}{2 a}\)
\(K=W=\frac{1}{2} m v^{2}\)

Potential Energy
The energy possessed by a body by virtue of its position or configuration is called the potential energy (J). When the position of an object or body or system changes, the potential energy changes and it converts into another form of energy.

A body kept at certain height has potential energy because when it is released, it provides motion to the body. The ability of a body kept at certain height of doing work by providing motion to the body is potential energy. When the body is released it acquires velocity. This means that it gains kinetic energy.

The gravitational force on a ball having mass m is mg. Let us raise the ball to a height h. Then, the work done by the external agent is mgh. This work gets stored as potential energy. So, the potential energy is (u).

U = W = mgh

Work-Energy Theorem Statement:
According to a work-energy theorem, the amount of work done by the net force acting on a body is equal to the change in kinetic energy. If the body is in motion. Mathematically, we write it as:

W = K– Ki – ΔK
Suppose a particle of mass m moving with a velocity u at any instant . Let a force \(\vec{F}\) be acting on the body due to which its  eleocity is \(\vec{u}\) after travelling a distance \vec{S} and the acceleration produced is a, then according to Newton’s third equation of motion;
\(\begin{aligned} v^{2} &=u^{2}+2 \vec{a} \vec{s} \\ \text { or } \quad v^{2}-u^{2} &=2 \vec{a} \vec{s} \end{aligned}\)

Multiplying the above equation with \frac{1}{2} m.

Hence, it is clear that the work done by the net force applied on a body is equal to the change in the kinetic energies of the body in two different situations. This is a work-energy theorem. Hence, it is clear that:
(i) \(W>0 \text { then }\left(K_{f}-K_{i}\right)>0 \text { or } K_{f}>K_{i}\) i.e., if work done is positive then final kinetic energy is more than initial kinetic energy.
(ii) \(W<0 \text { then }\left(K_{f}-K_{i}\right)<0 \text { or } K_{f}<K_{i}\) i.e., if work done is negative then final kinetic energy is less than initial kinetic energy.

Work energy theorem by calculus method :
Let the force (F) is applied on a body then the body is displaced at dx in the direction of force :
∴ Small work done = force x displacement
dW = Fdx
for total work done

Potential Energy of a Spring
An idealized spring exerts a force that is proportional to the amount of deformation of the spring from its equilibrium length; exerted in a direction opposite to the deformation. Pulling the spring to a greater length causes it to exert a force that brings the spring back towards its equilibrium length. The amount of force can be determined by multiplying the spring constant of the spring by the amount of stretch.

This opposite force is called ‘restoring force’. Restoring force is a force that gives rise to an equilibrium in a physical system. The force law for spring is called the Hooke’s law. If the restoring force is \(\vec{F}_{s}\) and displacement is \vec{x} then;
\(\begin{array}{l}{\vec{F}_{s} \propto-\vec{x}} \\ {\text { or } \quad \vec{F}_{s}=-k \vec{x}}\end{array}\) ……………….(1)

Where k is a constant which is called the force constant of the spring.

The negative sign in Eq. (1) shows that the restoring force i.e., the spring force (Fig.) is opposite to the displacement \(\overrightarrow{(x)}\). The unit of k from equation (1) is Nm-1

In equilibrium state the value of restoring force is calculated by external force. Hence, if external force is \(\vec{F}\) then;
\(F=-\vec{F}_{s}\)
∴ The form of eq. (1) will be as follows :
\(\begin{array}{l}{\vec{F}=k \vec{x}} \\ {\vec{F} \propto \vec{x}} \\ {F \propto x}\end{array}\) ………………….(2)
Hence, graph between F and × will be a straight line as shown in fig. 

Hence, from graph work done in giving a displacement × to the spring is;

Law of Conservation of Mechanical Energy
‘The mechanical energy of any body or a system i. e., the sum of its kinetic energy and potential energy is constant in the presence of conservative forces.’ This is called the law of conservation of mechanical energy.

We clearly know that when conservative forces one applied on a system, its configuration changes and the kinetic energy of this system changes by AK. According to the definition of conservative force there should be a change equal in magnitude in the potential energy of the system but in opposite direction. Due to which the sum of both the changes becomes zero, i. e.,
\(\begin{aligned} \Delta U &=-\Delta K \\ \text { or } \quad \Delta U+\Delta K &=0 \\ \text { or } \quad U+K &=\text { constant } \end{aligned}\)
= constant ………………………(1)

Hence, in the presence of the conservative forces, the change in the kinetic energy (K) is equal and opposite to the change in the potential energy (U). Due to this the sum of the kinetic energy and potential energy is always constant.

This law is not true for non-conservative forces like-frictional force because some part of the non-conservative forces changes into the sound, heat, light and other types of energies.

Some Examples of Law of Conservation of Mechanical Energy:
Example: Free Falling Body : Fig 5.7 shows a free falling body of mass m in the effect of gravitational force (conservative force). The height of point A from the earth’s surface is h; and the velocity of the object at A is zero (u = 0). Calculate the total mechanical energy of the system at A

When the body falls freely for a height × and reaches B then

∴ Total mechanical energy B

It is clear from the eqs (2), (3) and (4) that;
EA = EB = EC = mgh

Clearly, as the body falls, it’s potential energy decrease uniformly such that it’s total mechanical energy remains constant (mgh) at all point. Thus total mechanical energy is conserved during free fall of a body. Fig. shows the variation of KE and PE and the total mechanical energy (TME) with height

The relation between Linear Momentum and Kinetic Energy
We know that;

Eqs (1) and (2) show the relationship between linear momentum and kinetic energy. It is also clearly shown from these equations that without kinetic energy linear momentum is not possible and without linear momentum, kinetic energy is not possible.
Special Cases:

  1. If mass m is constant then \(p \propto \sqrt{K}\) i.e., for equal masses the object with more kinetic energy will have more linear momentum.
  2. If momentum p is constant then \(K \propto \frac{1}{m}\) i.e., for constant momentum of bodies, the one with minimum mass will have maximum kinetic energy.
  3. If kinetic energy is constant then linear momentum \(p \propto \sqrt{m}\) i. e., the body with maximum mass will have maximum linear momentum.

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