# what is conversion and conservation of energy

In physics energy can be defined as the ability or capacity to do work. The **unit** of work and energy are the same which is **Joule.**

## Forms of energy

(a). **Kinetic energy**:- this is the energy in moving body.

(b) **Chemical energy**: – energy in our food, motor battery, petrol is all chemical energy

(c). **Heat energy:** – heat from sun, car or steam engine is all heat energy.

(d). **electrical energy**: – This is the energy we use in running our television, refrigerators, and so on.

(e). **Light energy**: – Such energy from electric bulb which allows us to see.

(f) **Sound energy** such as that from speakers.

(g)** Nuclear energy**: – this is energy from the nucleus of an atom.

(h). **elastic energy**: – Energy stored in a spring.

**Conversion of energy from one form to another**

Energy can be transformed from one form to another, by means of suitable machines or apparatus. For instance energy can be converted to one another like this :

**Chemical energy from fuel => mechanical energy = > Electrical energy => sound , light or heat energy. or Solar energy => stored as chemical energy => light, sound or heat energy.**

There are many more ways energy can be converted to another form. Even the Telephone works with the principle of energy conversion, from sound to electrical and back to sound. The same conversion happens, when brake is applied to a moving car , the mechanical energy from the car is not lost , but is changed to heat and sound energy.

**The principle of conservation of energy.**

**It states that though energy can be changed from one form to another, but energy can neither be created nor destroyed during the transformation.**

** ****Types of energy**

We have only two types of energy . They are:-

**Kinetic energy**: – It is energy a body poses due its motion. Kinetic energy can simply be defined as energy at motion. Kinetic energy can be represented by symbol E_{k.}

**Potential energy**: – it is defined as energy at rest. If a body of mass M is raised to height h above the ground level is said to poses potential energy. The quantity of energy is given by:-

**E _{p }= mgh.**

**Potential energy = work done, that is E _{p} = force x distance**

**E _{p }= mg xh**

The potential energy in a stretched spiral spring is given by E_{p} =1/2ke^{2 }, where k = elastic constant, e = extension

Example: a body of mass **100kg** is released from a height of **200m**. With what energy does the body strike the ground? **(Take g = 10ms ^{-2})**

Solution, **E _{p} =mgh, where m = 100kg, g = 10m^{-2}, and h = 200m**

**E _{p} = 100 x 10 x x200 = 200000J = 200kJ**

**Example of body that poses kinetic energy are**

**A rolling ball, a moving car, a bullet fired from gun, wind or air in motion, an athlete running a race, flying plane e.t.c.**

Note that the more the speed the more the kinetic energy a body posses and also the greater the mass the more kinetic energy. For instance kinetic energy of a lorry moving with a certain speed will be more than a small car moving with the same speed, also cannon ball moving with a speed of 100kmh-^{1 }will knock down a mud wall more easily than it would if it was moving at 2kmh-^{1}

To obtain the expression for kinetic energy of a body of mass m, and speed v, let consider that the body is acted upon by a constant retarding force f, until body come to rest in a distance s. So the work done in bringing the body to rest from initial speed to final speed zero, is equal to the kinetic energy of the body when the speed is v, that is work done = force x distance = fS

**But S = average speed x time = 1/2vt**

**And f = ma = m xv/t, therefore work done = 1/2vt x mv/t = 1/2mv ^{2}**

**That is E _{K }=1/2mv^{2 }, where m = mass in kg, v = speed in ms-^{1 }**

**Example, A body of mass 30kg is running with a speed of 4ms- ^{1.} , what Is his kinetic energy**

**Solution, recall that E _{K} = 1/2mv^{2 }, therefore, E_{K} = ½ x 30 x 4 x 4 = 240J**

Example 2, a bullet of mass 40g is moving with a speed of 216 kmh-^{1 }calculate its kinetic energy.

**Solution, using E _{k} = 1/2mv^{2 }, where m = 40g, v = 216kmh^{-1 }, first convert all the units to SI units**

**40 = 40/1000n = 0.04kg, 216kmh- ^{1 }= 216 x 1000/ 60 x 60 = 60ms-^{1 }, therefore, kinetic energy**

**E _{K }= ½ x 60 x 60 x 0.04 = 72J.**

#### Conversion of kinetic energy to potential energy.

**Potential energy can be transformed to kinetic energy and vice versa, but in all case the total energy remain constant, that E _{K} = E_{P}_{ }= constant.**

The typical example of alternating potential and kinetic energy is that of simple pendulum

When the pendulum move from end to end, the energy of the system changes from potential to kinetic and vice versa, but at each stage the total energy remains constant or the same. From the diagram above, at C, the kinetic energy is at its maximum, but potential energy is zero, while at both A and B potential energy is at its maximum, and kinetic energy is zero

Therefore the total energy at C is equal to the kinetic energy of the system, while the total energy at A or B is equal to the potential energy of the system. Because energy can either be created or destroy, the total energy at C is total energy at A or C, that is

# mgh = 1/2 mv^{2}

Example 1. A stone of mass 0.5kg is thrown vertically upwards with a velocity of 10ms-^{2},

Find, ( a) the potential energy at the greatest height and the value of h. ( b) the kinetic energy on reaching the ground again. (Assume g = 10ms-^{2 }and neglect air resistance)

**Solution, recall that mgh = 1/2mv ^{2 }, where m = 0.5kg, v = 10ms-^{1 }, therefore**

**Mgh =1/2 x 0.5 x 10 ^{2 }= 25J**

**For h, taking mgh = 25, 0.5 x 10 xh = 25, 5h = 25, h = 25/5 = 5, there h = 5m**

(b) the kinetic energy on reaching the ground again will have the same value as 25J, since mechanical energy is the same provided no energy is lost to resistance and friction

Example 2. A mass of 2kg falls from rest through a height of 20m and comes to rest having penetrated a distance of 0.5m into the ground, calculate the average force exerted by the in bring the body to rest. ( take g = 10m-^{2})

**Solution, let u = velocity on reaching the ground, therefore**

**½ mv ^{2} = mgh, but m = 2kg, g = 10ms-^{2}, and h = 20m, therefore**

**½ mv2 = mgh = 2 x 10 x 20 = 400J,**

The kinetic energy is converted into work which is done in penetrating a distance S = 0.5, with an average force f that is ,** F x S = 400**

**0.5 x F = 400 ie F= 400/.o5 = 800 therefore**

** F = 800N.**

Example 3. A ball of mass 2kg falls from a height of 200m. Calculate its kinetic energy after falling a distance of 50m ( neglect air resistance, take g =10ms-2 )

**Solution , total energy at 50m = Ep = 2 x 10 x 50 = 1000J, but Ep = Ek at any point,**

**Therefore at 50m kinetic energy = 1000J = 1kJ**

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