In the previous section, we completed a discussion on work and energy'. In this section we will see power
We have seen some basics about power in our high school classes. The following link will give those details
High school physics - Power
It is recommended that the reader get a thorough understanding of those lessons before taking up our present discussion
1. Power is defined as the time rate at which work is done or energy is transferred
• 'time rate' means 'per unit time'
• Also note the terms:
♦ 'work is done' or
♦ 'energy is transferred'
2. So we can write:
■ In a process where work is done by a machine:
(eg: A motor of a lift used in high rise buildings)
• Power of the machine is the work done by it in unit time
■ In a process where energy is transferred:
(eq: An electric cooker which converts electrical energy into heat energy)
• Power of the appliance is the energy consumed by it in unit time
3. The unit of power is watt. It's symbol is W
• For example, a machine capable of giving 500 W power, may not give the same 500 joules of work every second
♦ During certain time intervals, it may be 540 J
♦ During certain other time intervals, it may be 480 J
2. In such cases, we take the average power
• It is done as follows:
(i) Turn on the machine and a stop-watch at the same instant
(ii) Record the total work 'W' done by the machine during a time interval 't'.
('t' may be any suitable time interval. Say 25 mins, 40 mins, 1 hr, 3 hr, etc.,)
(iii) Then the average power is given by: pav = W⁄t
• 1 hp = 746 W
• It is mainly used to express the power of automobiles, pumps etc.,
An elevator can lift a maximum load of 1800 kg. (This load includes the mass of the passengers + the mass of the elevator). It is carrying this maximum load and is moving up with a constant velocity of 2 ms-1. The frictional force opposing the motion is 4000 N. Calculate the power of the motor which lifts up the elevator [g = 10 ms-2]
Solution:
1. Power is 'work done in one second'
• Let us first find the work done in a suitable time interval say 10 s
2. 'Work done' is force × displacement
• The force here is the weight of the lift = mg = 1800 × 10 = 18000 N
• displacement is the distance traveled in 10 s
• Since the velocity is uniform, we can write:
displacement = velocity × time = 2 × 10 = 20 m
• So work done during the 10 s = 18000 × 20 = 360000 J
3. In addition to the above, work is done against friction also
• This work is given by:
frictional force × distance = 4000 × 20 = 80000 J
4. So total work done during the 10 s = 360000 + 80000 = 440000 J
5. So Power = work done in 1 s = 440000⁄10 = 44000 W = 44 kW
6. 1 hp = 746 W
• So 1 W = 1⁄746 hp
• So 44000 W = 44000 × 1⁄746 = 58.98 = 59 hp
Solved example 6.29
A pump at the ground floor of a building can pump up water to fill up a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground and the efficiency of the pump is 30%, how much electric power is consumed by the pump? [g = 9.8 ms-2]
Solution:
1. Total work done by the pump
= Work done in completely filling the tank
= Work done in lifting 30 m3 of water to an elevation of 40 m
= mgh
= mass of 30 m3 of water × 9.8 × 40
2. So our next task is to find the ' mass of 30 m3 of water'
• We have: Density = mass⁄volume
• So mass = volume × density = 30 × 1000 = 30000 kg
(∵ Density of water = 1000 kg m-3)
3. So the result in (1) becomes:
• Total work done by the pump = 30000 × 9.8 × 40 = 11760000 J
4. The efficiency is 30 %
• Let 'I' be the total electrical energy input during the 15 min
• Only 30% of I will be converted into useful work
• The rest will be lost as heat and sound
5. The 'useful work required' in our present case is 11760000 J
• So we can write:
I × 0.3 = 11760000 J
• Thus I = 39200000 J
6. This much electrical energy is consumed in 15 min
So power = $\mathbf\small{\frac{\text{work}}{\text{time}}=\frac{39200000}{15\times 60}=43555.55 \,\,\text{W}=43.55\,\,\text{kW}}$
Solved example 6.30
A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this can be converted into useful electrical energy, how large an area is needed to supply 8 kW. (b) Compare this area to that of the roof of a typical house.
Solution:
1. In this problem, the consumption is given in terms of power. Not in terms of energy
(In fact, consumption is indeed given in terms of power. But for solving numeric problems, we can convert it into energy using suitable time intervals)
• It is given that the family uses 8 kW of power
• We can say: The family uses 8000 J of energy every second
2. The available solar energy is also given in terms of power
• It is given that 1 m2 area receives a power of 200 W
• We can say: 1 m2 area receives 200 J of energy every second
3. Only 20% of this will be converted into electrical energy
• So we can say: 1 m2 area provide an energy of 200 × 0.2 = 40 J (every second)
4. But the requirement is 8000 J in every second
• This can be achieved by increasing the area
• So area required = Total energy required⁄Energy obtained per unit area = 8000⁄40 = 200 m2
5. A typical house has 9 rooms each of average size 3.6 × 3.6 m
• So area of the roof = 9 × 3.6 × 3.6 = 116.64 m2.
♦ This is more than 'half of 200'
• So more than half of the energy requirement can be met using solar energy
Solved example 6.31
A train moves with a constant velocity of 40 ms-1 on a level road. The friction and air resistance has a total magnitude of 3 × 104 N. What is the power of the engine?
Solution:
1. Power is 'work done in one second'
• Let us first find the work done in a suitable time interval say 10 s
2. 'Work done' is force × displacement
On a level road, no work is done by the engine 'for moving the train' (Details here)
3. But work is done against friction and air resistance
• This work is given by: force × distance = 30000 × (40 ×10) = 12000000 J
[∵ uniform velocity of 40 ms-1 for time10 s will give a distance of (40 ×10) m]
4. So total work done during the 10 s = 12000000 J
5. So Power = work done in 1 s = 12000000⁄10 = 1200000 W = 1200 kW
Solved example 6.32
A train of mass 2 × 105 kg is moving up a hill with a constant velocity of 20 ms-1. The inclination of the hill is [sin-1 (1⁄50)] to the horizontal. For this motion, the engine works at 8 × 105 W. What is the resistance to motion of the train? [g = 9.8 ms-2]
Solution:
1. The engine is working at 8 × 105 W
• That means, the engine is doing 8 × 105 J of work every second
2. Let us consider a suitable time interval say 10 s
• During this 10 s, the engine will do a work of (8 × 105 × 10) = 8 × 106 J
3. Let the train move from A to B during those 10 s
This is shown in the fig.6.63(a) below:
4. Let the level difference between A and B be 'h'
• Then in fig.6.63(b), the altitude BC of the right triangle = 'h' m
5. Distance AB = (velocity × time) = (20 × 10) = 200 m
• In the right triangle ABC, we have: sin θ = h⁄AB = h⁄200
• But sin θ = 1⁄50.
• So we get: h⁄200 = 1⁄50
• So h = 200⁄50 = 4 m.
6. Work done by the engine during those 10 s is the sum of the following two items:
(i) Work done for raising the train through 4 m
(ii) Work done against 'resistance to motion'
• Work done for raising the train through 4 m = mgh = 2 × 105 × 9.8 × 4 = 78.4 × 105 J.
7. So we can write:
78.4 × 105 + Work done against 'resistance to motion' = 8 × 106 J
• So we get:
Work done against 'resistance to motion' = (8 × 106 - 78.4 × 105) = 1.6 × 105 J
• So during those 10 s, 1.6 × 105 J of work is done against 'resistance to motion'
8. We have: Work done = force × distance
• So we can write:
1.6 × 105 = (force required to overcome 'resistance to motion') × distance
⇒ 1.6 × 105 = (force required to overcome 'resistance to motion') × AB
⇒ 1.6 × 105 = (force required to overcome 'resistance to motion') × 200 m
⇒ force required to overcome 'resistance to motion' = 800 N
By Newton's third law,
9. force required to overcome 'resistance to motion' = 'resistance to motion'
• Thus the required answer is 800 N
We have seen some basics about power in our high school classes. The following link will give those details
High school physics - Power
It is recommended that the reader get a thorough understanding of those lessons before taking up our present discussion
1. Power is defined as the time rate at which work is done or energy is transferred
• 'time rate' means 'per unit time'
• Also note the terms:
♦ 'work is done' or
♦ 'energy is transferred'
2. So we can write:
■ In a process where work is done by a machine:
(eg: A motor of a lift used in high rise buildings)
• Power of the machine is the work done by it in unit time
■ In a process where energy is transferred:
(eq: An electric cooker which converts electrical energy into heat energy)
• Power of the appliance is the energy consumed by it in unit time
3. The unit of power is watt. It's symbol is W
Average power
1. In some cases, the 'work done' or 'energy consumed' may not be uniform
1. In some cases, the 'work done' or 'energy consumed' may not be uniform
♦ During certain time intervals, it may be 540 J
♦ During certain other time intervals, it may be 480 J
2. In such cases, we take the average power
• It is done as follows:
(i) Turn on the machine and a stop-watch at the same instant
(ii) Record the total work 'W' done by the machine during a time interval 't'.
('t' may be any suitable time interval. Say 25 mins, 40 mins, 1 hr, 3 hr, etc.,)
(iii) Then the average power is given by: pav = W⁄t
■ Horse power is another unit for power. It's symbol is hp
• It is mainly used to express the power of automobiles, pumps etc.,
Solved example 6.28
Solution:
1. Power is 'work done in one second'
• Let us first find the work done in a suitable time interval say 10 s
2. 'Work done' is force × displacement
• The force here is the weight of the lift = mg = 1800 × 10 = 18000 N
• displacement is the distance traveled in 10 s
• Since the velocity is uniform, we can write:
displacement = velocity × time = 2 × 10 = 20 m
• So work done during the 10 s = 18000 × 20 = 360000 J
3. In addition to the above, work is done against friction also
• This work is given by:
frictional force × distance = 4000 × 20 = 80000 J
4. So total work done during the 10 s = 360000 + 80000 = 440000 J
5. So Power = work done in 1 s = 440000⁄10 = 44000 W = 44 kW
6. 1 hp = 746 W
• So 1 W = 1⁄746 hp
• So 44000 W = 44000 × 1⁄746 = 58.98 = 59 hp
Solved example 6.29
A pump at the ground floor of a building can pump up water to fill up a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground and the efficiency of the pump is 30%, how much electric power is consumed by the pump? [g = 9.8 ms-2]
Solution:
1. Total work done by the pump
= Work done in completely filling the tank
= Work done in lifting 30 m3 of water to an elevation of 40 m
= mgh
= mass of 30 m3 of water × 9.8 × 40
2. So our next task is to find the ' mass of 30 m3 of water'
• We have: Density = mass⁄volume
• So mass = volume × density = 30 × 1000 = 30000 kg
(∵ Density of water = 1000 kg m-3)
3. So the result in (1) becomes:
• Total work done by the pump = 30000 × 9.8 × 40 = 11760000 J
4. The efficiency is 30 %
• Let 'I' be the total electrical energy input during the 15 min
• Only 30% of I will be converted into useful work
• The rest will be lost as heat and sound
5. The 'useful work required' in our present case is 11760000 J
• So we can write:
I × 0.3 = 11760000 J
• Thus I = 39200000 J
6. This much electrical energy is consumed in 15 min
So power = $\mathbf\small{\frac{\text{work}}{\text{time}}=\frac{39200000}{15\times 60}=43555.55 \,\,\text{W}=43.55\,\,\text{kW}}$
Solved example 6.30
A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this can be converted into useful electrical energy, how large an area is needed to supply 8 kW. (b) Compare this area to that of the roof of a typical house.
Solution:
1. In this problem, the consumption is given in terms of power. Not in terms of energy
(In fact, consumption is indeed given in terms of power. But for solving numeric problems, we can convert it into energy using suitable time intervals)
• It is given that the family uses 8 kW of power
• We can say: The family uses 8000 J of energy every second
2. The available solar energy is also given in terms of power
• It is given that 1 m2 area receives a power of 200 W
• We can say: 1 m2 area receives 200 J of energy every second
3. Only 20% of this will be converted into electrical energy
• So we can say: 1 m2 area provide an energy of 200 × 0.2 = 40 J (every second)
4. But the requirement is 8000 J in every second
• This can be achieved by increasing the area
• So area required = Total energy required⁄Energy obtained per unit area = 8000⁄40 = 200 m2
5. A typical house has 9 rooms each of average size 3.6 × 3.6 m
• So area of the roof = 9 × 3.6 × 3.6 = 116.64 m2.
♦ This is more than 'half of 200'
• So more than half of the energy requirement can be met using solar energy
Solved example 6.31
A train moves with a constant velocity of 40 ms-1 on a level road. The friction and air resistance has a total magnitude of 3 × 104 N. What is the power of the engine?
Solution:
1. Power is 'work done in one second'
• Let us first find the work done in a suitable time interval say 10 s
2. 'Work done' is force × displacement
On a level road, no work is done by the engine 'for moving the train' (Details here)
3. But work is done against friction and air resistance
• This work is given by: force × distance = 30000 × (40 ×10) = 12000000 J
[∵ uniform velocity of 40 ms-1 for time10 s will give a distance of (40 ×10) m]
4. So total work done during the 10 s = 12000000 J
5. So Power = work done in 1 s = 12000000⁄10 = 1200000 W = 1200 kW
Solved example 6.32
A train of mass 2 × 105 kg is moving up a hill with a constant velocity of 20 ms-1. The inclination of the hill is [sin-1 (1⁄50)] to the horizontal. For this motion, the engine works at 8 × 105 W. What is the resistance to motion of the train? [g = 9.8 ms-2]
Solution:
1. The engine is working at 8 × 105 W
• That means, the engine is doing 8 × 105 J of work every second
2. Let us consider a suitable time interval say 10 s
• During this 10 s, the engine will do a work of (8 × 105 × 10) = 8 × 106 J
3. Let the train move from A to B during those 10 s
This is shown in the fig.6.63(a) below:
 |
| Fig.6.63 |
• Then in fig.6.63(b), the altitude BC of the right triangle = 'h' m
5. Distance AB = (velocity × time) = (20 × 10) = 200 m
• In the right triangle ABC, we have: sin θ = h⁄AB = h⁄200
• But sin θ = 1⁄50.
• So we get: h⁄200 = 1⁄50
• So h = 200⁄50 = 4 m.
6. Work done by the engine during those 10 s is the sum of the following two items:
(i) Work done for raising the train through 4 m
(ii) Work done against 'resistance to motion'
• Work done for raising the train through 4 m = mgh = 2 × 105 × 9.8 × 4 = 78.4 × 105 J.
7. So we can write:
78.4 × 105 + Work done against 'resistance to motion' = 8 × 106 J
• So we get:
Work done against 'resistance to motion' = (8 × 106 - 78.4 × 105) = 1.6 × 105 J
• So during those 10 s, 1.6 × 105 J of work is done against 'resistance to motion'
8. We have: Work done = force × distance
• So we can write:
1.6 × 105 = (force required to overcome 'resistance to motion') × distance
⇒ 1.6 × 105 = (force required to overcome 'resistance to motion') × AB
⇒ 1.6 × 105 = (force required to overcome 'resistance to motion') × 200 m
⇒ force required to overcome 'resistance to motion' = 800 N
By Newton's third law,
9. force required to overcome 'resistance to motion' = 'resistance to motion'
• Thus the required answer is 800 N
In the next section we will see collisions
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