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.
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
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.
• 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
• 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.
• If the power of an object is 1 kW, it consumes or delivers 1000 joules in every second.
• 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
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
• 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
In the next section, we will see Sound and Wave motion.
■ 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.
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
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.
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
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
Now we will see a solved example
Solved example 4.9Two 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
• 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
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
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
In the next section, we will see Sound and Wave motion.
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