What is electro-motive force (e.m.f.)?

As you would remember from this post on complete circuit a power supply is needed in order to move charges in the circuit.

Thus if we consider that an electric current is a flow of positive charges according to the conventional definition of current, for the positive charges to move from the positive terminal to the negative terminal of the power supply they need energy. It is this energy that is supplied by the power supply to the positive charges. The positive charges can thus use this energy to move around the circuit. It is this energy that the positive charges possesses that is called electrical energy.

Fig 1

Fig 1  shows a closed circuit. If the circuit is closed then an electric current will flow as indicated from the positive terminal of the power supply to the negative terminal of the power supply.

Fig 2

In fig 2 you can see 1 c of charge leaving the positive terminal of the power supply. Now for this 1 C of charge the power supply will provide a certain amount of energy. 

Suppose that for every one coulomb of charge that leaves the power supply the power supply provides 10 J of energy. Then the electromotive force (e.m.f.)  of the power supply will be 10 Volt (V).

Hence if the e.m.f. of a dry cell is 1.5 V it means that when the dry cell is in a circuit for every 1 C of charge then the power supply will provide 1.5 J of energy.

Hence we can define the electromotive force as the energy that the power supply provide to a unit charge to move it around the circuit.

Hence if the a charge Q is moved around a circuit and the power supply supplied W amount of energy, the the e.m.f. E can be calculated as shown:

E  = W/Q

Example

If 4 C of charge  needs 10 J of energy is provided by the power supply. Calculate the e.m.f. of the power supply?

E = W/Q

= 10 /4

2.5 J

How to calculate the energy of a photon?

You would remember from a previous post that a photon is a packet of energy. Hence all electromagnetic radiations are emitted in term of photons.

 

We also saw that the energy of a photon is given by the equation

E = hf

where h is the Planck’s constant and f is the frequency of the electromagnetic radiation’s photon.

Fig 1

From the equation we can thus deduce that a photon’s energy is directly proportional to its frequency.

Hence the higher the frequency of a photon the greater the energy it has.

From fig 1 we can deduce that since radio wave has the least frequency then the photons of radio waves has the least energy. And as shown in fig 1 as we move from left to right across the electromagnetic spectrum the frequency increases. This means that the energy of the photon also increases. Photons of gamma rays would thus be the most energetic.

Example

Calculate the energy of a photon of red light of wavelength 650 nm if the speed of light is 3.0 x 10⁸ m/s and Planck’s constant = 06.63 x 10^-34 m² kg s^-1.

Now if you remember the frequency f, the wavelength λ and the speed  c of all electromagnetic radiations are related by the equation

c = fλ

hence making f the subject of formula

f =c/λ

Calculating f

f = 3.0 x 10⁸ /650 x 10^-9

= 4.6 x 10¹⁴ Hz

Calculating the energy of the photon using the equation E = hf

Energy E = 6.63 x 10^-34 * 4.6 x 10¹⁴

                     =3.0 x10^-19 J

You would remember in a previous post that we discussed about the use of the electron volt eV as unit of energy in quantum mechanics.

We are now going to see how to determine the energy of the photon in eV

E = 3.0 x10^-19 J

= 3.0 x10^-19 /1.6 x10^-19

= 1.9 eV

As we have seen in the previous post working with eV is much better.

What is the electron volt (eV)?

The electron volt is a unit of energy like the joule (J).

However in quantum mechanics if the joule is used, then very small numbers would be obtained. These small numbers are very clumsy to use as a result a better unit to use is the electron volt.  

Fig 1

As you can see in fig 1, we have two metal plates that are connected to a power supply of e.m.f. 1 V. As a result a potential difference of 1 V will be set up be set up between the plates. An electric field will be set up between the plates as a result of the potential difference.

Now if a stationary electron is released from the negatively charged plate it will experience an acceleration due to the electric field and as a result will move towards the positively charged plate with increasing velocity.

When it would reached the positively charged plate it would have a certain velocity.

The Kinetic energy gained by the electron can be calculated by the following equation:

Kinetic energy gained E_k = Charge on particle * Potential difference between plates.

E_k = Q * V

Hence it can be deduce that if the charge is an electron then the kinetic energy that is gained by the electron is

Kinetic energy gained E_k = Charge on electron (elementary charge) * 1 V

   = 1 eV

Hence the eV is the energy that is gained by an electron or any particle of charge -e or +e when accelerated by a potential difference of 1 V 

If you want to know how many joule there is in 1 eV then it is as shown below

Kinetic energy gained E_k = Charge on electron (elementary charge) * 1 V

= 1.6 x 10^-19* 1

= 1.6 x10^-19 J

We can thus say that 1 eV is equivalent to 1.6 x 10 ^–19 J

As we have said above in quantum mechanics it is better to use the the eV because if the joule is used we would be dealing with small numbers which would make the works and calculations difficult.

What is the kilowatt-hour?

The kilowatt-hour is the unit in which usage of electricity is measured in the home.

You would remember that the SI unit of energy is the Joule (J)  as you have seen in  work done, gravitational potential energy or kinetic energy.

Now as you know the rating of home appliances is its power and the unit of power is the Watt (W).

Now if a kettle with a rating of 1000 W is used for two hours, then the amount of energy used will be calculated as shown:

Power = Energy used / time taken

Energy used = power * time taken  = 1000 * (2 * 3600)

                    = 7200000 J

                     = 7.2 MW (Megawatt)

However as you can see such calculation is difficult and beyond someone that is not a Physicist.

Hence in order to make matter more simple we use the Kilowatt-hour.

If we use the example above we would need to use the equation below:

Energy used in kilowatt-hour = power of appliance in Kilowatt  x time in hour

Hence the energy used = 2 kW x 2 h

= 4  kilowatt-hour (kWh)

As you can see it is quite simple to determine the the energy used in kilowatt-hour.

What is a photon?

A photon also known as a corpuscle or a quantum is the basic unit of electromagnetic radiation.

You will remember that the electromagnetic spectrum is made up of different radiations. These radiations are waves and as such they are energy being moved from one place to another.

The photon is simply a small packet of energy and a beam of electromagnetic radiation is just a stream of these photons travelling in a particular direction.

So if the photon is a packet of energy then you should be able to determine the energy of a photon of electromagnetic radiation.

The following equation is used:

Energy of a photon = Planck’s constant * frequency of electromagnetic radiation

E = h * f

Planck’s constant = 6.63 x 10^-34 m² kg s^-1

If you want to calculate the energy of a photon read this post .

Force, work, power and energy

Newton’s first law of motion

Newton’s second law of motion 

Weight of an object

Work done

Kinetic energy

Gravitational Potential energy

Gravitational potential energy to kinetic energy and vice versa

Power

Gravitational potential energy to kinetic energy and vice versa.

 

As we have seen in earlier post an object can have either gravitational potential energy or kinetic energy. However it is always possible for the object to have both kinetic energy and gravitational potential energy. Think of a plane flying at a certain height above the ground.

It is going to have kinetic energy due to its speed and gravitational potential energy due to its height.

Now what happens to a body that is either falling toward the ground or rising to a certain height. 

As you can see in fig 1 the object is rising and as a result its height is also rising, hence its potential energy is also increasing. Its kinetic energy is however decreasing. (see Principle of conservation of energy).

However the object in fig 2 is falling towards the ground. Since its height is decreasing, its gravitational potential energy must also be decreasing and its kinetic energy increasing.

Example 1

In this example a ball is released from rest from a height h.

If the ball is initially at rest then

v = 0  ms^-1 hence E_k = 0 J

However since the object is at a height h the E_p = mgh

When the ball is released it fall under the effect of the force of gravity, as a result it will accelerate downward and as a result the velocity of the object will increase  while the height of the object decreases.

Hence as the velocity decreases the kinetic energy decreases while the gravitational potential energy decreases as the height decreases.

However at all time    E_k +  E_p = Total energy and total energy is constant.

As the object reaches the ground the height becomes zero so does the gravitational potential energy while the kinetic energy reaches its maximum value. At this point all the gravitational potential energy would have been converted to kinetic energy.

When it reaches the ground h = 0 m hence E_p = 0 J

While E_k = 0.5 mv²

Example 1

A man of mass 64 kg jumps from a bridge 25 m high into a river.

(a) Calculate the gravitational potential energy of the man when he is on the bridge.

(b) What is his speed of entry into the water.

Now the man is on the bridge at a height of 25 m. It means that he has gravitational potential energy.

(a)  Gravitational potential energy E_p = mgh

                                                                = 64*9.81*25

                                                                = 15696J

                                                                 = 1.5 *10⁴J                                        

(b) When the person jumps the gravitational potential energy decreases as his height decreases. However as the person fall to the ground his speed increases and as a result the kinetic energy is completely converted to kinetic energy.

Hence what he reaches the river all the gravitational potential energy has been converted to kinetic energy.

Kinetic energy at surface of river =  1.5 *10⁴J

E_k = 0.5 mv²

1.5 *10⁴= 0.5 *64*v²

v = (1.5 *10⁴/0.5/64)^0.5

    = 21.65 m s^-1

=22 m s^-1

It is now time for a question. I will give you an answer to do. I will give the answer when some of you have given the answers.

Good luck.

A girl of mass 50 kg is trying to jump over a bar. She ran at a speed of and leaves the ground and successfully jumped over the bar.

(a) Calculate the kinetic energy that she has when she is running.

(b) Deduce the gravitational potential energy of the girl when she is at her maximum height.

(c) Calculate the height of the bar.

Kinetic energy

Kinetic energy is one if the eight forms of energy. It is simply the energy that a body possesses due to its motion.

The kinetic energy of a body is simply calculated using the equation

Kinetic energy Ek = ½ mv²

Where m is the mass of the mass of the object in kg
v is the velocity of the object in ms^-1

As you can see the kinetic energy of the object depends on both the velocity and the mass of the object. So it is possible for an elephant with a large mass to have the same kinetic energy as a small object moving at a high velocity.

Now let us see an example where the kinetic energy of an object is calculated.

Example 1

An elephant of mass _{3.00 x 10}³ kg is moving at a velocity of 3.00 ms^-1. What is the kinetic energy of the elephant?

So let us recall that the kinetic energy of the elephant is calculated using the equation

Ek = ½ mv²

Hence the kinetic energy Ek = ½ mv²
= ½(_{3.00* 10}³)*(3.00)²
= 13500 J
= _{1.35 *10}⁴ J

Now you can be given the kinetic energy and be asked to calculate the mass or the velocity. Let us look at a second example.

Example 2

An object has a mass of 4.0 kg and a kinetic energy of 16 J. Determine the velocity of the object.

Ek = ½ mv²
16 = ½ (4.0)*v²

Making v the subject of formula

v² = (16 *2)/4.0
v = (16*2/4.0)^½
v = 2.8 ms^-1

Now to test your newly acquired you can do the following questions

1. Calculate the kinetic energy possessed by an aeroplane of mass _{3.4 * 10}⁵ kg flying at 110 ms.
2. If a body has _{3.4 x 10}⁵ J of kinetic energy and mass _{3.2* 10}² kg, what is its velocity?

What is power?

Power is the rate of doing work.

From the definition we can deduce the following equation

Power = Work done /Time taken

The unit of power is the Watt (symbol W) or the Joule/second (J/s)

Example 1

A boy pushes a box and as a result does 120 J of work in 10 s. What is the power developed by the boy?

Power = Work done / time taken

= 120 /10

= 12 W or J/s

There is another definition for power that is often used. It is

Power is the rate of dissipation of energy or the rate of change of energy conversion.

Power = Energy dissipated / Time taken

Example 2

A girl climbs a staircase gaining 500 J of gravitational potential energy in 10 s.

What is the power developed by the girl?

Power = Energy conversion / Time taken

= 500 / 10

= 50 W or J/s

Example 3

During the boiling of some water 4000 J of heat energy is dissipated in the kettle’s heater in a time of 8 s. What is the power of the heater?

Power = Energy dissipated/Time taken

= 4000/8

= 500 W or J/s

It is now time to do some questions. It will give these questions after a few of you have supplied your answers.

1. A trains of mass 50000 kg accelerated form rest and reaches a velocity of 50 ms^-1 in 60 s.

(a) Calculate the kinetic energy gained by the train.

(b) Calculate the power of the train engine.

2. A lamp is rated 80 W. If it is switched on for two hours, how much light energy is dissipated.

Work done

Work done is a very important concept in Physics as it is used in fields, in deriving gravitational potential energy and kinetic energy, etc.

So what is work done?

Let us look at the diagram above. A force F acts on the box at A. During the time the force is acting, the box moves in the direction of the force. When the application of the force stopped the box has moved a distance d.

Using the definition

Work done is the product of the force acting on an object and the distance moved by the object in the direction of the force.

We can deduce that the equation to calculate work done is

Work done = Force * distance moved in the direction of the force

Work done = F *d

Let us have a look at an example.

Example

A man pulls a table by exerting a force of 100 N on it moving it by a distance of 3.9 m.

Calculate the work done by the man?

Work done = Force * distance moved in the direction of the force

Work done = F *d

= 100 * 3.9

= 3900 J

If you have understood the concept do the question below.

Question

A braking force of 2.3*10⁴ N is applied to a car and as a result the car stops in a distance of 23 m. Calculate the work done in stopping the train.

Good luck.