# Energy

Without force, work cannot be done. But, if we do not have energy in our body, we cannot apply force. Thus, energy is defined as the capacity or ability to do work.
The SI unit of energy is joule.
The amount of energy which can move a body of weight 1 Newton through 1 m distance is called 1 joule of energy.
Energy is the amount of work an object can do. Electrical energy is used by the bulb to glow and so the energy is used by a fan to rotate.

### Forms of Energy

There are many forms of energy. They are;
1. Mechanical energy
2. Heat energy
3. Light energy
4. Chemical energy
5. Electrical energy
6. Magnetic energy
7. Sound energy
8. Nuclear energy

### Types of Energy

In dynamics, there are two types of energies; Kinetic Energy and Potential Energy.

## Kinetic Energy

Kinetic energy is the energy possessed by a body by the virtue of its motion. It is measured by the amount of work which the body can do against any force before coming to rest. A bullet fired from a gun and body falling from a certain height possess kinetic energy.

Let a body of mass m have the velocity u. Let F be the force force applied on the body to resist its motion. If a is the retardation produced due to the force F and s is the distance covered by the body before coming to rest, then $0=u^2-2as$ $as=\frac{u^2}{2}$ Now, K.E of the body = Work done against the force before coming to rest $\text{K.E.}=Fs$ $=(ma)s$ $=m(as)$ $=m\frac{u^2}{2}$ $=\frac{1}{2}mu^2$

### Work Energy Theorem

It states that​
Work done by a force on a body is equal to the change in kinetic energy of the body.
Consider a body of mass m moving with initial velocity u. Let a constant force F is applied on the body till its velocity becomes v. Let s be the displacement of the body from its initial point to the point where its velocity becomes v. Let a be the acceleration produced in the body by the applied force.​

​$\text{For θ=0°, work done is,}$ $W=Fs$ $W=mas\text{__(1) [F=ma]}$ $\text{From linear equation of motion,}$ $v²=u²+2as$ $2as=v²-u²$ $as=\frac{v²-u²}{2}\text{__(2)}$ From (1) and (2), $W=m\left(\frac{v²-u²}{2}\right)$ $W=\frac{1}{2}mv²-\frac{1}{2}mu²$ $∴\text{Work done}=\text{Change in kinetic energy}$ This is work energy theorem.

## Potential Energy

Potential energy is the energy produced by a body by virtue of its position or state. It is measured by the amount of work which the body can do in changing from its actual position to some standard position. A weight lifted to a certain height possesses potential energy. Greater the height ascended by the body, greater will be the potential energy in magnitude.

Consider a body of mass m held stationery at a height h above the surface of the earth. If the body is allowed to fall freely, the force of gravity acts on the body. If the body reaches the ground, then, $\text{Work done}=\text{Force}×\text{Displacement}$ $=\text{Weight}×\text{Height}$ $=mgh$ The unit of energy is same as work since it is measured in terms of the amount of work done. Hence, the SI unit of energy is Joule (J) and CGS unit is Erg. Energy is also a scalar quantity like work.

## Transformation of Energy

Transformation of energy is the change of one form of energy into another form. There are many examples for transformation of energy. A battery converts chemical energy into electrical energy. A candle converts chemical energy into heat and light energy. A bulb converts electrical energy into heat and light energy.
Transformation of energy also takes place in our body. We eat food and the food gets digested. When we respirate, the chemical energy of the digested food changes into heat energy. This heat energy maintains warmness in our body making us able to do different types of work.

## Principle of Conservation of Energy

​Statement:
Energy can neither be created nor can be destroyed but it can be converted from one form to another.
In other words,
The total energy of an isolated system always remains constant.
For example, the mechanical energy (sum of kinetic energy and potential energy) of a body falling freely under gravity always remains constant.

Illustration

​Consider a body of mass m is falling freely under gravity from height h.

At point A,
Velocity of the body is 0. $(K.E.)_A=\frac{1}{2}mv_A²$ $(K.E.)_A=\frac{1}{2}m×0$ $(K.E.)_A=0$ and, $(P.E.)_A=mgh$ $∴\text{Total Mechanical Energy at A}=(K.E.)_A+(P.E.)_A$ $=0+mgh$ $=mgh$
At point B,
the body is (h-x) height above from the ground. $∴(P.E.)_B=mg(h-x)$ $=mgh-mgx$ velocity at point B, $v_B²=v_A²+2gx$ $v_B²=2gx$ $∴(K.E)_B=\frac{1}{2}mv_B²$ $=\frac{1}{2}m(2gx)$ $=mgx$ $∴\text{Total Mechanical Energy at B}=(K.E.)_B+(P.E.)_B$ $=mgh-mgx+mgx$ $=mgh$
At point C,
​The body is at height equal to 0. $∴(P.E.)_C=0$ and, velocity at C, $v_C²=v_A²+2gh$ $v_C²=2gh$ Then, $(K.E.)_C=\frac{1}{2}mv_C²$ $=\frac{1}{2}m×2gh$ $=mgh$ $∴\text{Total Mechanical Energy at C}=(K.E.)_C+(P.E.)_C$ $=mgh+0$ $=mgh$ From the above discussion, we can conclude that the total mechanical energy of the body at the points A, B and C remains constant which verifies the principle of conservation of energy.

Q. A particle is dropped from a height h. At what height is its K.E. equal to 1/4 of its P.E.?

Let the mass of the particle be m and let AB=h. Let C be the point at which the K.E. is equal to 1/4 of the P.E.

If BC=x, then AC=h-x. $\text{P.E. at C}=mgx$ If v is the velocity of the particle at C, then, $v^2=2g(h-x)$ $\text{Thus, K.E. at C}=\frac{1}{2}mv^2$ $=\frac{1}{2}m \cdot 2g(h-x)$ $=mg(h-x)$ $\text{From Question,}$ $\text{K.E. at C}=\frac{1}{4} \text{P.E. at C}$ $mg(h-x)=\frac{1}{4}mgx$ $4h-4x=x$ $4h=5x$ $x=\frac{4}{5}h$ Thus, K.E. is equal to 1/4 of P.E. at height 4h/5.

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