Chapter 10.3 - Electric Power

In the previous section we completed a discussion on the 'lighting effect of electric current'. In this section, we will see 'electric power'.

• We have seen in earlier classes that, power is: Work done in unit time
• But 'work done' is same as 'energy consumed'. Both have the same unit joule
• So another way of defining power is: Energy consumed in unit time
We have seen those details here.

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• Now, in our present discussion, consider any electric appliance. 
• We know that, it will consume electric energy. By consuming energy, it does work. 
■ How much work does it produce? OR, How much energy does it consume? 
We cannot give an exact answer. Because, it depends on time. 
• If it works for a long time, we say: It does a lot of work for us
    ♦ Same as: It consumes a lot of energy from us
• If it works for a short time, we say: It does only a little work for us 
    ♦ Same as: It consumes only a little energy from us
■ So time is an important factor.
• 'Work done per unit time' is what matters more.
• Some heavy appliances can do more work per unit time. 
    ♦ Unit time can be taken as one second, one minute, one hour etc.,
    ♦ Usually we take  'one second'
• Some lighter appliances can do only a less work per unit time. 

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Based on the above, we can derive an equation:
• Let an appliance consume electric energy for 't' seconds
• As a result, let it produce a work of 'H' joules
■ Then the power of that appliance is given by: P = ^H⁄_{t
• The unit of power is Watt}

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We will apply this to a heating coil. We will write it in steps:
1. Let us give electrical energy to that heating coil, for 't' seconds.
• Then, based on joules law, it will produces a work (in the form of heat) given by H = I²Rt
[Here, 'heat energy produced' can be considered as 'work'. Because, that heat can be used to produce steam, which can turn a small turbine for example] 
2. So in 1 second, it produces a work of ^I2Rt⁄_t 
• 't' in numerator and denominator cancels each other. Thus we can write: 
Eq.10.4: P = I²R

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• The equation P = I²R gives us the relation between power and the quantities: [current, resistance]
• Can we relate power to voltage? Let us try:
• From ohm's law, we have: V = IR (Details here)
• From this we get: I = ^V⁄_R
• So we can use (^V⁄_R) instead of I
• Thus we get: P = (^V⁄_R)²R ⟹ P = (^V2⁄_R²)R =  ^V2⁄_R
So we can write:
Eq.10.5: P = ^V2⁄_R 

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Another derivation:
• From ohm's law, we have: V = IR
• From this we get: R = ^V⁄_I
• So we can use (^V⁄_I) instead of R
• Thus we get: P = I²(^V⁄_I) ⟹ P = VI
So we can write:
Eq.10.6: P = VI

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Now we will see a solved example:
Solved example 10.3:
A heating appliance has a resistance of 115 Ω. If 2 A current flows through it, what is the power of the appliance?
Solution:
1. Given: R = 115 Ω, I = 2 A
2. We want a relation that will connect the power (P) to: [R, I]
• So we will use Eq.10.4: P = I²R
3. Substituting the values, we get: P = 2² × 115 = 460 W

Solved example 10.4
An appliance of power 540 W is used in a branch circuit. If the voltage is 230 V, What is the amperage?
Solution:
1. Given: P = 540 W, V = 230 volts
2. We are asked to find the amperage (I). So we want a relation that will connect the power (P) to: [V, I]
• We will use Eq.10.6: P = VI
3. Substituting the values, we get: 540 = 230 × I ⟹ I = ⁵⁴⁰⁄₂₃₀ = 2.3478 A
• The next whole number should be taken as the amperage. So we get:
Amperage = 3 A

Solved example 10.5
A current of 0.4 A flows through an electric bulb working at 230 V. What is the power of the bulb?
Solution:
• Given: When the bulb is connected to a circuit and a voltage of 230 V is applied, a current of 0.4 A flows through it.
• If we apply less than 230 V, only a low current will flow. This will be less than the 'current specified by the manufacturer' of the bulb. 
• Because of such a low current, the filament will not get heated to the required temperature, and so we will get only a dim light.
• Thus, 0.4 A is the 'rated current' of the bulb. For this current, 230 V is essential.
• This current and voltage are related to the resistance of the bulb by the relation V = IR (Details here)
• For solving this problem however, we do not need to write the above details
1. Given: I = 0.4 A, V = 230 volts
2. We are asked to find the power. So we want a relation that will connect the power (P) to: [V, I]
• We will use Eq.10.6: P = VI
3. Substituting the values, we get: P = 230 × 0.4  = 92 W

Solved example 10.6
A device of resistance 690 Ω is working at 230 V. What is the power of the device?
Solution:
1. Given: R = 690 Ω, V = 230 volts
2. We are asked to find the power. So we want a relation that will connect the power (P) to: [R, V]
• So we will use Eq.10.5: P = ^V2⁄_R
3. Substituting the values, we get: P = ²³⁰²⁄₆₉₀ = 76.67 W

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Now we will see the 'changes in power' when changes occur in other factors: 
• resistance • current • voltage
1. First we will see the 'changes in power' when 'change in resistance and current' occur:
Consider a bulb filament which is broken at a point along it's length. 
• The filament is broken into two unusable pieces. 
2. If we join the two pieces together, current can flow again. 
• But when the joining is done, the length of the filament will decrease. 
3. 'Decrease in the length of a conductor' will decrease the resistance
• Because resistance is directly proportional to length (Details here)
4. Now, if the resistance decrease, the current will increase. 
• This is clear from the ohm's law: V = IR (Details here)
An example:
■ If the voltage is constant, calculate the current when resistance is halved
Solution: 
(i) Let the constant voltage be V
(ii) Let the initial current and resistance be I₁ and R₁ respectively 
• Then we can write: V = (I₁ × R₁)
(iii) Let the new current and resistance be I₂ and R₂ respectively 
• Then we can write: V = (I₂ × R₂)
(iv) Since voltage is constant at V, we can equate the results in (ii) and (iii):
(I₁ × R₁) = (I₂ × R₂)
(v) Given that R₂ = ^R₁⁄₂
• Substituting this in (iv), we get:
(I₁ × R₁) = (I₂ × ^R₁⁄₂) ⟹ (I₁) = (^I₂⁄₂) ⟹ I₂ = 2I₁
• We can write: When resistance is halved, current becomes double  
5. We also have the Eq.10.4: P = I²R, which relates power to current
■ From this equation, we see that, when current increases, power increases   
So we can write:
■ When resistance decreases, current increases and so, power also increases
6. Next we will see the 'changes in power' when 'change in voltage' occurs:
• We have the Eq.10.5: P = ^V2⁄_R, which relates power to voltage
■ From this equation, we see that, when voltage increases, power increases

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Solved example 10.7
An instrument is marked 150 W, 230 V. If the voltage is lowered to 110 V, what will be it's power?
Solution:
1. Given: P = 150 W, V = 230 volts
2. The manufacturer knows that, if the customer supplies a voltage of 230 V, that instrument will do a work of 150 joules in one second (∵ power is marked as 150 W)
• How is the manufacturer so sure about it?
• Ans: Because he has fitted a fixed resistance (R) in that instrument
3. So we want an equation which relates power to [voltage, resistance]
• We can use Eq.10.5: P = ^V2⁄_R
• Substituting the values, we get: 150 = ²³⁰²⁄_R ⟹ R = ²³⁰²⁄₁₅₀ = 352.67 Ω 
4. So the manufacturer has fitted a resistance of 352.67 Ω in that instrument. This resistance value will not change.
• Now, if we supply only 110 volts (instead of 230), the instrument will not give us a work of 150 joules in one second 
• It will be less than 150. How much will that be?
• Ans: we again use Eq.10.5: P = ^V2⁄_R
• We get: P = ¹¹⁰²⁄_352.67 = 34.31 W
5. Now we know the reason why some instruments slow down when voltage is low.

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In the next section we will see a few more solved examples.

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Chapter 10 - Effects of Electric Current

In the previous section we completed a discussion on the 'relation between electricity and magnetism'. In this chapter, we will see the 'effects of electric current'.

Let us first see how an electric current can produce a 'heating effect'. We will do an activity:
1. In the fig.10.1 below, A and B are two beakers of 200 mL capacity.

Fig.10.1

• Each beaker contains 100 mL of water.
• The conductor PQ in the beaker A is a nichrome wire
• The conductor RS in the beaker B is a copper wire
• Both the conductors have the same length and diameter
2. Trial 1:
Measure the temperature of the water in both the beakers and note it down. Now we can start the trials:
• Turn the switch on
• Note down the current shown by the ammeter reading
• Allow the current to flow through the circuit for three or four minutes 
• Turn off the switch and immediately measure the temperature of water in each beaker. 
    ♦ Let them be t_A and t_B
• We will see that t_A is larger than t_B
3. So the observations are over. Let us analyse them:
• Beakers A and B are connected in series. So the same current (indicated by the ammeter) will be flowing through both nichrome and copper
• Both nichrome and copper pieces are of the same size. Because they have the same length and diameter
• Even when current, lengths and diameters are the same, nichrome produced greater heat. Why is that so?
Ans: Nichrome offers greater resistance to the flow of electricity. That is., electricity cannot flow easily through nichrome. So most of the electrical energy will be converted into heat energy.
4. But the opposite happens in the case of copper
• Copper offers lesser resistance to the flow of electricity. That is., electricity can flow easily through copper. So only a lesser electrical energy will be converted into heat energy.
• Since in this trial, current, lengths and diameters of both nichrome and copper are the same, we can say that, resistance to flow of electricity depends on the material with which the conductor is made. We discussed it in a previous chapter here.
■ So we can write: Greater the resistance (R), greater is the heat energy produced

5. Trial 2:
• Allow the water in the beakers to cool down to room temperature. 

• Remove the beaker B from the circuit. Now there is only the nichrome conductor in beaker A in the circuit. This is shown in fig.10.1(b) above

• Measure the temperature of water in beaker A and note it down.
• Turn on the switch and pass current for five minutes
• Note down the current. Let it be I₁
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t₁
6. Allow the water to cool down back to room temperature
• Turn on the switch and pass current for seven minutes
• The current must be the same I₁ as before. Ensure this by the ammeter reading
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t₂
• We will see that, t₂ is greater than t₁
■ So we can write: Greater the 'time duration (t) for which current is passed through a conductor', greater is the heat energy produced

7. Trial 3:
• Allow the water to cool down to room temperature. Measure this temperature and note it down
• Turn the switch on and allow the current to pass for three minutes
• Note down the current I₁
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t₁
8. Allow the water to cool down back to room temperature
• Turn on the switch and pass current for the same three minutes 
• The current I₁ this time must be greater than I₁. 
    ♦ Ensure this by the ammeter reading. The current can be increased by adjusting the rheostat
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t₂
We will see that t₂ is greater than t₁
■ So we can write: Greater the 'current (I) which is passed through a conductor', greater is the heat energy produced

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From the above activity, we can write:
The heat energy produced in a conductor due to the flow of current through it, depends on three items:
(i) The resistance (R) of the conductor
(ii) The time duration (t) for which the current is passed through the conductor
(iii) The intensity of the current (I)
• The English scientist James Prescott Joule discovered the relation connecting those three factors. The relation is known as Joule's Law:
■ The heat generated in a current carrying conductor is the product of the square of the current (I) in the conductor, the resistance (R) of the conductor and the time (t) of the flow of current

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In simple terms, we can say:
The heat will be equal to the product of three items:
(i) Square of the current (I²)
(ii) Resistance (R)
(iii) time (t)
• So if H is the heat generated, then:
Eq.10.1: H = I²Rt 
• I is measured in ampere, R in ohm and t in seconds

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Let us see some interesting cases:

■ Heat of H₁ joules is generated in a conductor when a current I₁ passes through it for t seconds. How much heat will be generated if the current is doubled and is passed for the same t seconds?
Solution:
• We have: H = I²Rt 
• Case 1: H₁ = (I₁)²Rt 
• Case 2: Let H₂ be the heat generated when a current I₂ passes for the same t seconds
• Then we can write:
H₂ = (I₂)²Rt
• Taking ratios, we get: ^H₁⁄_H2= ^(I₁)2Rt⁄_(I2)²Rt = ^(I₁)2⁄_(I2)² = [^(I₁)⁄_(I2)]² 
But given that  I₂ = 2I₁
• So we get: ^H₁⁄_H2 = [^(I₁)⁄_(2I1)]²  = [⁽¹⁾⁄₍₂₎]²  = [¹⁄₄]
⟹ ^H₁⁄_H2 = ¹⁄₄ ⟹ H₂ = 4H₁
• So we can write: The new heat will be four times the original heat

■ Heat of H₁ joules is generated in a conductor when a current I₁ passes through it for t seconds. How much heat will be generated if the current is halved and is passed for the same t seconds?
Solution:
• We have: H = I²Rt 
• Case 1: H₁ = (I₁)²Rt 
• Case 2: Let H₂ be the heat generated when a current I₂ passes for t seconds
• Then we can write:
H₂ = (I₂)²Rt
• Taking ratios, we get: ^H₁⁄_H2= ^(I₁)2Rt⁄_(I2)²Rt = ^(I₁)2⁄_(I2)² = [^(I₁)⁄_(I2)]² 
But given that  I₂ = 0.5I₁
• So we get: ^H₁⁄_H2 = [^(I₁)⁄_(0.5I1)]²  = [⁽¹⁾⁄_(0.5)]²  = [⁽¹⁾⁄_(5/10)]²  = [⁽¹⁰⁾⁄₍₅₎]²  = [2]²  = 4
⟹ ^H₁⁄_H2 = 4 ⟹ H₂ = ^H₁⁄₄ 
• So we can write: The new heat will be one fourth of the original heat

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• The equation H = I²Rt gives us the relation between heat and the quantities: [current, resistance, time]
• Can we relate heat to voltage? Let us try:
• From ohm's law, we have: V = IR (Details here)
• From this we get: I = ^V⁄_R
• So we can use (^V⁄_R) instead of I
• Thus we get: H = (^V⁄_R)²Rt ⟹ H = (^V2⁄_R²)Rt =  (^V2⁄_R)t = ^V2t⁄_R
So we can write:
Eq.10.2: H = ^V2t⁄_R 

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Another derivation:
• From ohm's law, we have: V = IR
• From this we get: R = ^V⁄_I
• So we can use (^V⁄_I) instead of R
• Thus we get: H = I²(^V⁄_I)t ⟹ H = VIt
So we can write:
Eq.10.3: H = VIt

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Now we will see a solved example:
Solved example 10.1:
A bulb of resistance 920 Ω works on 230 V supply. Calculate the quantity of heat generated in 3 minutes
Solution:
1. Given: R = 920 Ω, V = 230 volts, t = 3 minutes = (3 × 60) = 180 seconds 
2. We want a relation that will connect the heat (H) to: [R, V and t]
• So we will use Eq.10.2: H = ^V2t⁄_R 
3. Substituting the values, we get: H = ^(230)2×180⁄₉₂₀ = 10350 J
Another method:
1. We have the basic equation: H = I²Rt 
2. But the current I is not given. We can calculate it using the Ohm's law: V = IR
• So we get: I = ^V⁄_R = ²³⁰⁄₉₂₀ = ¹⁄₄ = 0.25 A   
3. Thus we get: H = I²Rt = H = (0.25)² × 920 × 180 = 10350 J
One more method:
1. Here we will try to use Eq.10.3: H = VIt
• We have calculated I in the previous method. So we can use it
2. Substituting the values, we get: H = 230 × 0.25 × 180 = 10350 J

Solved example 10.2
An electric iron works on 230 V. A current of 3 A flows through it for half an hour. Calculate the amount of heat energy generated
Solution:
1. Given: V = 230 volts, I = 3 A, t = 30 minutes = (30 × 60) = 1800 seconds
2. We want a relation that will connect the heat (H) to: [V, I and t]
• So we will use Eq.10.3: H = VIt
3. Substituting the values, we get: H = 230 × 3 × 1800 = 1242000 J = 1242 kJ

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■ So we have seen a special property of electricity:
• It will produce heat when it passes through a conductor. 
• We also saw how to calculate the 'quantity of heat' produced. 
• In the next section we will see some practical applications of this property.

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Chapter 4.4 - Relation between Energy and Power

In the previous section we saw the Law of conservation of energy. In this section we will see the basic details about Power.

■ Consider two labourers doing a certain work. 
• The first labourer has to load 50 bricks onto a platform at a height h. 
• The second labourer has to load another 50 bricks onto the same platform at height h.
• So the work to be done are same for both the labourers.
■ Now, out of the two labourers one is stronger than the other. 
• But based on what we have learned so far, there is no way to give any 'special consideration' for him. Because, both will be doing the same joules of work. 
■ So we introduce a new term power. 
• We have to calculate the power of each labourer separately. 
• Then we can give 'special consideration' for the labourer who shows greater power. 

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So we want to know how the power is calculated. Let us see:
1. Let one of the labourer start to load the bricks. 
• The moment when he starts to do the work, a stop watch should be started. 
2. Note the time when all 50 bricks are loaded. 
• Thus we get the time t1 required by the first labourer to do the work. 
3. Repeat the process for the second labourer. Let the time required by him be t2
4. We know that work done by both is the same. Let this work be W joules.
Then we get:
• Power of the first labourer = ^W⁄_t1
• Power of the second labourer = ^W⁄_t2

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So, to calculate the power, we are dividing the work by time. 

■ That means, power is the work done in unit time. 

We can write an equation:

Eq.4.3:
Power p = ^W⁄_t
• Note that, time t is in the denominator. So, if t decreases, power increases. 
• In the above example, the stronger labourer will most probably complete the work in a lesser time. 
• Thus we would be able to say that he has greater power.

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In the same way,

■ Consider a distance d
• A car having greater power travels this distance in time t1
• An ordinary car travels this same distance in time t2
• t1 will be less than t2
■ Consider two heaps of sand. Each having the same volume v
• An earth mover having greater power will carry away it's heap in t1
• An ordinary earth mover will carry away it's heap in time t2
• t1 will be less than t2

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Next we want a unit for power. 

■ We have seen that p = ^W⁄_t . 

• The unit of work is joules. 

• The unit of time is seconds. 

• So the unit of p is joules per second. It is written as J s^-1. 

• There is a special name for this J s^-1. It is watt. It's short form is W. This name is given in honour of the British scientist James Watt, who invented the Steam engine. 

■ We can say that power of an engine is 1 W, if it delivers 1 joule of work in every second. 
• We know that, to deliver work, an object has to consume energy. 
    ♦ The energy may be the chemical energy in petrol, coal etc., or 
    ♦ Electrical energy, heat energy etc., 
■ If an object consumes 1 joule of energy in every second, then also we say that, the power of that object is 1 W.

■ For expressing larger powers, we use larger forms of watt:

• 1 kilowatt = 1000 watts. In short form, it is: 1 kW = 1000 W
• If the power of an object is 1 kW, it consumes or delivers 1000 joules in every second.

Average power

• Consider an engine having a power of 500 W. Let it work for 2 hours. 
• We cannot expect it to give 500 joules every second. There may be variations. 
• It may give greater than 500 for some time duration, and less than 500 for some other time duration.
• In such cases, we first calculate the total work done in the duration of 2 hours. 
    ♦ Then we divide this total work by 2 × 3600 = 7200 seconds. 
• The result that we get is called the Average power. 
■ So average power can be defined as: ^{Total work}⁄_{Total time}

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Now we will see a solved example

Solved example 4.9
Two workers: A and B carry a weight of 350 N each to The second floor of a building, which is 6 m above the ground level. A takes 40 seconds and B takes 55 seconds to do the task. What is the power expended by each worker?
Solution:
1. Work done = Force × displacement = Weight × height
Weight = mg. It is given as 350 N
2. So work done = mgh = 350 × 6 = 2100 J
Both A and B do this same work. But the time taken is different
3. We have: Power p =  ^W⁄_t.
• So power expended by A = ²¹⁰⁰⁄₄₀ = 52.5 J s^-1 = 52.5 watts    
• Power expended by B = ²¹⁰⁰⁄₅₀ = 42 J s^-1 = 42 watts

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• So far in this section, we have been discussing about power. We saw the unit of power also. 
• Now we are going to revisit energy. In fact we are going to discuss about a unit of energy. It is called 'kilowatt hour'. It's symbol is kW h. Let us analyse this kW h:
1. We know that W is a unit for power. It is one joule work in one second. That is., one joule per second
2. So kiloWatt (kW) is 1000 joules in one second. That is 1000 joules per second
3. 'per second' means that the time (in seconds) is in the denominator
• That is., 1 kW = ^{1000 joules}⁄_{1 second}.
4. In kW h, we are not saying 'per hour'. That means the time (in hours) is in the numerator
5. So we can write: 1 kW h = [(^{1000 joules}⁄_{1 second} )×1 hour] = [(^{1000 joules}⁄_{1 second} )×3600 seconds]
• The seconds in numerator and denominator cancels out. What we get is:
1 kW h = 1000 × 3600 joules = 3600000 joules = 3.6 × 10⁶ J   
• So, kWh is a unit of energy. It is not a unit of power
■ 1 kW h is the total energy delivered by a machine in one hour, if it delivers 1000 joules in every second
OR
■ 1 kW h is the total energy consumed by a machine in one hour, if it consumes 1000 joules in every second

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So 1 kWh is a large quantity of energy. It is 36 followed by five zeroes. This large quantity of energy is used as '1 unit' for measuring electrical energy consumed in our homes, schools, offices etc.,

Let us see an example:

1. In an electricity bill, the following two details are written:

(a) Previous reading: 13541 in the electric meter

(b) Present reading: 13892

2. Calculate the difference: 

It is equal to: 13892 - 13541 = 351

3. So 351 kWh of electrical energy was consumed during the time between the following two points:

(a) The instant when previous reading was taken

(b) The instant when present reading is taken

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Now we will see some solved examples

Solved example 4.10

An electric bulb of 40 W is used for 5 hours per day. Calculate the ‘units’ of energy consumed in one day by the bulb.

Solution:

1. 'One W' is 'one joule' consumed every second. 

2. So 40 W is 40 joules consumed every second

3. So in one hour it will consume 40 × 60 × 60 = 144000 joules

4. So in 5 hours it will consume 144000 × 5 = 720000 joules

5. 1 unit = 1 kWh = 3600000 joules

6. So 'No. of units' contained inside 720000 joules =  ⁷²⁰⁰⁰⁰⁄_3600000 = 0.2

7. Thus we can write: When a 40 W bulb is used for 5 hours, 0.2 units of energy will be used up.

Solved example 4.11

Find the energy in kWh consumed in 6 hours by three devices of power 300 W each.

Solution:

Power of one device = 300 W

1. So it will consume 300 joules in every second

2. That means, it will consume 300 × 3600 = 1080000 joules in one hour 

3. So in 6 hours it will consume 6 × 1080000 = 6480000 joules

4. So three such devices will consume 3 × 6480000 = 19440000 joules in 6 hours

• One kWh = 3600000 joules.

5. So 19440000 joules = ^19440000⁄_3600000 = 5.4 kWh = 5.4 units

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In the next section, we will see Sound and Wave motion. 

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Chapter 4 - Work, Power and Energy

In the previous section we saw mass, weight, thrust and pressure. In this section we will see work, power and energy.
■ Consider some of our day to day activities. We walk, run, play, read, think. We engage in many such activities. 
• For performing those activities, we need energy. Some activities like running and playing requires more energy. 
• We get the required energy from the food we eat.
■ Consider animals. They also perform various activities. Like running, hunting, hiding from enemies etc.,  Also we humans make them do activities like pulling vehicles, carrying loads etc., 
• Animals also get the required energy from the food that they eat
■ Consider machines. They perform activities like pumping water, pulling trains, lifting heavy loads etc., 
• They get their energy from the fuels like petrol, coal, diesel etc.,

To define energy from a scientific point of view, we must first understand about another concept 'work'. In day to day life we regularly come across the term work. Let us see some examples:
• Two laborers worked hard to load the bricks into the truck
• Much work is required to arrange all the newly arrived books in the library
• A person is holding a heavy luggage on his shoulders. He may have to stand in that position for a long time. He is applying upward force through his shoulders. He may even get exhausted. But we would not say that he is 'working' to keep the luggage on his shoulders.
• Talking with friends for a long time about some academic topics may be considered as 'work'. It involves expenditure of energy too.
• At the same time, taking with friends about a movie may not be considered as work
■ Thus, even in common day do day conversations, we give a difference between work and energy. 
• In day to day conversations, we would not mind if they are interchanged. 
• From a scientific point of view, such interchanging is not allowed.

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■ In science, work is done only if two conditions are satisfied: 

• A force should act on an object
• The object must get displaced due to the force
Let us see some examples:
1. Push a pebble lying on the ground. The pebble moves through a distance
• A pushing force was applied on the object (the object here is pebble)
• The object got displaced
■ So work is done
2. A girl pulls a trolley. The trolley moves through a distance
• A pulling force was applied on the object (the object here is trolley)
• The object got displaced
■ So work is done
3. Lift a book through a height. For lifting the book, a force must be applied. The book rises up.
• A lifting force was applied on the object (the object here is book)
• The object got displaced
■ So work is done
4. Push a large rock on the ground. The rock does not move.
• A pushing force was applied on the object (the object here is rock)
• The object did not get displaced
■ So in this case, work is not done

Work done by constant force

We are now able to identify those situations in which a work is done. What we need next, is a method to calculate the amount of work done.
Let a constant force, F act on an object. Let the object be displaced through a distance, s in the direction of the force. This is shown in fig. 4.1 below:

Fig.4.1

Let W be the work done.
Then W = F × s
Thus, work done by a force is equal to the product of 
• The force and
• The displacement produced by the force
■ Next we need a unit for the force. 
• Unit for force is newton 
• Unit for distance is metre.
■ So unit for work is newton-metre. In short form, it is N m
• This 'N m' has another name: joule. In short form, it is J
• Let a force of 1 N move an object through a distance of 1 m. Then the amount of work done is obtained as follows:
W = Fs = 1 × 1 = 1N m = 1 joule
• Thus we can write:
1 N m or 1 joule of work is done on an object, if a force of 1 N moves it through a distance of 1 m along the line of action of the force.

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The unit 'joule' for work is named after James Prescott Joule. He was a British scientist who made significant contributions in the field of electricity and thermodynamics. 

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Now we will see a solved example:

Solved example 4.1

A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force. If the force acts on the object all through the displacement, calculate the work done.

Solution:

We have: W = F×s = 5 × 2 = 10 N m or 10 J

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When we defined work, we specified an important point. That is:

■ The displacement should be along the line of action of the force. 

Now consider the situation in fig.4.2 below:
1. A force F is acting on a body. The body is being displaced. The displacement is along the line of the force.

Fig.4.2

2. But the displacement is in the opposite direction of the force.
■ Let us see an example for such a situation:
• A body is moving with a uniform velocity u
• A retarding force F acts on it in a direction opposite to the direction of motion
• As a result of this retarding force, the body comes to rest. That is., final velocity v = 0 
• In this situation, how will we define force?
• We can use the same principle. But note that the displacement is opposite to the direction of the force. We know that displacement has both magnitude and direction. So it's direction should be taken as negative. This is to indicate that, it is opposite to the direction of the force.
So we get: W = F × (-s) = -Fs
■ Thus, the work can be either positive or negative
Another example:
1. Lift an object upwards. When we lift an object we are applying a force on the object
• The direction of movement of the object is same as the direction of the force
• So the work done is positive
2. When the object is lifted up, another force is also acting on the object. It is the gravitational force
• This force acts in a direction opposite to the direction of movement of the object
• So work done by the gravitational force is negative

Solved example 3.2
A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage. 
Solution:
1. Work done W = F×s
2. Here displacement s = 1.5 m
3. We need to find the force F. 
4. We know that the luggage which has a mass of 15 kg is acted upon by a force. This force is the 'Gravitational force' F_g exerted by the earth towards it's centre. 
5. We know that this force is equal to the weight mg. Where m is the mass and g the acceleration due to gravity.
So gravitational force of the earth = W = mg = 15 × 10 =150 N
6. To lift the object, we need to apply an equal and opposite force. That is., we need to apply 150 N in the upward direction. The displacement of 1.5 m is in the same direction of the lifting force. So we can take (+1.5) m
Thus work done = F×s = 150 ×1.5 = 225 J 

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In the next section, we will see Kinetic energy. 

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