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.
• 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
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:
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.
1. A force F is acting on a body. The body is being displaced. The displacement is along the line of the force.
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' Fg 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
In the next section, we will see Kinetic 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.
■ 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 |
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.
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.
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
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 |
■ 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' Fg 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
In the next section, we will see Kinetic energy.
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