Good morning, students! Welcome to today's science class. I am so happy to see all of you here, ready to learn something new and interesting. Today, we are going to study a very important chapter from your science textbook — Chapter 10: Work and Energy. This is a chapter that connects the concepts of motion that we studied earlier with the idea of energy, which is fundamental to understanding how everything in this world works. So, let's begin our journey into the world of work and energy.
Now, students, before we start with the formal definitions, let me ask you something. Have you ever thought about why we feel tired after playing cricket or after running around in the playground? Or why do we need to eat food every day? The answer lies in the concept of energy. All living beings need food to survive, and the energy from food helps us perform various activities like playing, singing, reading, writing, thinking, jumping, cycling, and running. Animals also need energy to perform their activities — they jump, run, fight enemies, find food, and sometimes they even help us by carrying loads, pulling carts, or ploughing fields. Machines too need energy to work. Think about the various machines you have seen — the fan that cools you in summer, the mixer grinder in your kitchen, the car that takes you from one place to another. What do all these machines need to work? They need fuel like petrol, diesel, or electricity, don't they? All these fuels are sources of energy.
Now, students, let me tell you about someone very close to my heart — my niece Kamali. She is preparing for her examinations and spends a lot of time studying. She reads books, draws diagrams, organises her thoughts, collects question papers, attends classes, discusses problems with her friends, and performs experiments. She expends a lot of energy on these activities. In common parlance, we say she is working hard. But here's an interesting point — all this hard work may involve very little work if we go by the scientific definition of work. Isn't that surprising? Let me explain what I mean.
Imagine you are working hard to push a huge rock. You put in all your effort, you strain yourself, you get completely exhausted. But the rock does not move even a little bit. Have you done any work on the rock in the scientific sense? The answer is no, because there is no displacement of the rock. Now, consider another situation. You stand still for a few minutes with a heavy load on your head. You get tired, you have exerted yourself, and you have spent quite a bit of your energy. Are you doing work on the load in the scientific sense? Again, the answer is no, because the load has not moved. However, when you climb up the steps of a staircase and reach the second floor of a building, or when you climb up a tall tree, then according to the scientific definition, these activities involve a lot of work because there is displacement involved.
So, students, you can see that there is a big difference between the way we use the term work in our daily life and the way scientists define work. In our day-to-day life, we consider any useful physical or mental labour as work. But in science, work has a very specific meaning. Let me make this clear with some more examples.
## 10.1 Work
So, what exactly is work in science? Let us consider a few situations. First, imagine you push a pebble lying on a surface. The pebble moves through a distance. You exerted a force on the pebble, and the pebble got displaced. In this situation, work is done. Now, imagine a girl pulling a trolley. The trolley moves through a distance. The girl has exerted a force on the trolley, and it is displaced. Therefore, work is done. Now, think about lifting a book through a height. To do this, you must apply a force. The book rises up. There is a force applied on the book, and the book has moved. Hence, work is done.
Now, students, look carefully at these three situations. What do they have in common? In each case, two conditions are satisfied: first, a force is acting on an object, and second, the object is displaced. These are the two conditions that need to be satisfied for work to be done in the scientific sense. If any one of these conditions does not exist, work is not done. This is the scientific definition of work.
Let me give you a few more examples to make this crystal clear. A bullock is pulling a cart. The cart moves. There is a force on the cart, and the cart has moved. Do you think that work is done in this situation? Yes, definitely! Now, let me ask you to think about some situations from your daily life. Can you list some situations where work is done? Try to think about it. When you push a shopping cart in a supermarket, work is done. When you lift your school bag, work is done. When you pedal your bicycle, work is done. In each of these situations, there is a force acting on an object, and the object is displaced.
Now, students, I want you to think about situations when the object is not displaced in spite of a force acting on it. Can you think of such situations? One example is when you push a wall — you apply force, but the wall does not move. Another example is when you hold a heavy object without moving it. In these situations, work is not done because there is no displacement. Now, can you think of situations when an object gets displaced in the absence of a force acting on it? This might seem tricky, but think about it. When an object is moving due to its own inertia, there may be no force acting on it at that moment, but it is still displaced. For example, when you throw a ball upwards, at the highest point, the velocity becomes zero for a moment, but the ball is still in the air. Actually, in most real situations, forces are involved, but in ideal physics problems, we sometimes consider situations where an object keeps moving due to its inertia.
### 10.1.3 Work Done by a Constant Force
Now, students, let us learn how to calculate work mathematically. We shall first consider the case when the force is acting in the direction of displacement. Let a constant force, denoted by F, act on an object. Let the object be displaced through a distance, denoted by s, in the direction of the force. Let W be the work done. We define work to be equal to the product of the force and displacement. So, Work done equals force multiplied by displacement. In mathematical form, we write this as W equals F times s. This is equation number 10.1 in your textbook.
Thus, work done by a force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force. Work has only magnitude and no direction. This means work is a scalar quantity, not a vector quantity.
Now, students, what is the unit of work? In equation 10.1, if F equals 1 newton and s equals 1 metre, then the work done by the force will be 1 newton metre. We give this unit a special name — it is called joule, named after James Prescott Joule, a famous British physicist. So, 1 joule is the amount of work done on an object when a force of 1 newton displaces it by 1 metre along the line of action of the force. Remember, 1 joule is written as 1 J.
Now, let me ask you a very important question. What is the work done when the force on the object is zero? If there is no force, then even if the object moves, the work done will be zero because F is zero. So, W equals zero times s equals zero. What would be the work done when the displacement of the object is zero? If the object does not move, then no matter how much force is applied, the work done will be zero because s is zero. So, W equals F times zero equals zero. This is very important, students — work is zero if either the force is zero or the displacement is zero.
Now, let us look at Example 10.1 in your textbook. A force of 5 newton is acting on an object. The object is displaced through 2 metres in the direction of the force. If the force acts on the object all through the displacement, then work done is 5 newton multiplied by 2 metre, which equals 10 newton metre or 10 joule. Simple, isn't it?
Now, students, there is something more we need to learn. Consider a situation in which an object is moving with a uniform velocity along a particular direction. Now, a retarding force, denoted by F, is applied in the opposite direction. That is, the angle between the two directions is 180 degrees. Let the object stop after a displacement s. In such a situation, the work done by the force is taken as negative and denoted by the minus sign. The work done by the force is F multiplied by minus s, which is minus F times s. So, work done can be either positive or negative.
Let me explain this more clearly. Work done is negative when the force acts opposite to the direction of displacement. For example, when you throw a ball upwards, gravity acts downwards while the ball is moving upwards. So, the force of gravity is acting opposite to the direction of displacement, and therefore the work done by gravity is negative. On the other hand, work done is positive when the force is in the direction of displacement. For example, when you drop a ball, gravity acts downwards and the ball also moves downwards, so the force of gravity is in the direction of displacement, and therefore the work done by gravity is positive.
Let us do a small activity to understand this better. Lift an object up. Work is done by the force exerted by you on the object. The object moves upwards. The force you exerted is in the direction of displacement. However, there is also the force of gravity acting on the object, which is downwards while the object is moving upwards. So, which one of these forces is doing positive work? The force you apply is doing positive work because it is in the direction of displacement. Which one is doing negative work? The force of gravity is doing negative work because it acts opposite to the direction of displacement. Give reasons for your answer. This is exactly what we just discussed.
Now, students, let us look at Example 10.2. A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. We need to calculate the work done by him on the luggage. The mass of the luggage is 15 kg, and the displacement is 1.5 m. In such situations, the work done by the force is taken as positive. The force applied by the porter is equal to the weight of the luggage, which is mg. So, work done W equals F times s equals mg times s. Substituting the values, we get W equals 15 kg times 10 m per second squared times 1.5 m. This equals 225 kg metre squared per second squared, which is 225 newton metre or 225 joule. So, the work done is 225 joule.
Now, students, let me ask you a question. Why did we take g as 10 m/s²? In many problems, for simplicity, we take g equal to 10 m/s² instead of 9.8 m/s². This makes calculations easier, and the answer will be very close to the actual value. In some problems, however, the textbook might give the value of g explicitly, and you should use that value.
Now, let us answer the questions from section 10.1.
Question 1: When do we say that work is done? We say that work is done when a force acts on an object and the object is displaced in the direction of the force.
Question 2: Write an expression for the work done when a force is acting on the object in the direction of its displacement. The expression is W equals F times s, where F is the force and s is the displacement.
Question 3: Define 1 joule of work. 1 joule is the amount of work done on an object when a force of 1 newton displaces it by 1 metre along the line of action of the force.
Question 4: A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field? Here, force F equals 140 N, and displacement s equals 15 m. So, work done W equals F times s equals 140 N times 15 m equals 2100 J. So, the work done is 2100 joule.
Now, students, before we move on to the next section, let me quickly recap what we have learned about work. Work is done when a force acts on an object and the object is displaced. Work is calculated as the product of force and displacement. The unit of work is joule. Work can be positive or negative depending on whether the force is in the same direction as the displacement or opposite to it.
## 10.2 Energy
Now, students, let us move on to a new but related concept — energy. Life is impossible without energy. The demand for energy is ever increasing. Where do we get energy from? The Sun is the biggest natural source of energy to us. Many of our energy sources are derived from the Sun. For example, the food we eat comes from plants, which get their energy from the Sun through photosynthesis. The fossil fuels like coal and petroleum were formed from ancient plants and animals that derived their energy from the Sun millions of years ago. We can also get energy from the nuclei of atoms, which is nuclear energy. We can get energy from the interior of the Earth, which is geothermal energy. We can also get energy from the tides, which is tidal energy. Can you think of other sources of energy? Yes, there is wind energy, solar energy, hydroelectric energy, and so on.
Now, students, the word energy is very often used in our daily life, but in science we give it a definite and precise meaning. Let us consider some examples to understand what energy means in science.
When a fast-moving cricket ball hits a stationary wicket, the wicket is thrown away. The cricket ball has done work on the wicket. Similarly, an object when raised to a certain height gets the capability to do work. You must have seen that when a raised hammer falls on a nail placed on a piece of wood, it drives the nail into the wood. The hammer has done work on the nail. We have also observed children winding a toy, such as a toy car, and when the toy is placed on the floor, it starts moving. When a balloon is filled with air and we press it, we notice a change in its shape. As long as we press it gently, it can come back to its original shape when the force is withdrawn. However, if we press the balloon hard, it can even explode, producing a blasting sound. In all these examples, the objects acquire, through different means, the capability of doing work. An object having a capability to do work is said to possess energy. The object which does the work loses energy, and the object on which the work is done gains energy.
So, students, energy is the capability of an object to do work. How does an object with energy do work? An object that possesses energy can exert a force on another object. When this happens, energy is transferred from the former to the latter. The second object may move as it receives energy and therefore do some work. Thus, the first object had a capacity to do work. This implies that any object that possesses energy can do work.
The energy possessed by an object is thus measured in terms of its capacity of doing work. The unit of energy is, therefore, the same as that of work, that is, joule. 1 joule is the energy required to do 1 joule of work. Sometimes a larger unit of energy called kilojoule is used. 1 kJ equals 1000 J.
Now, students, let me tell you about a very famous scientist — James Prescott Joule. He was an outstanding British physicist who lived from 1818 to 1889. He is best known for his research in electricity and thermodynamics. Amongst other things, he formulated a law for the heating effect of electric current. He also verified experimentally the law of conservation of energy and discovered the value of the mechanical equivalent of heat. The unit of energy and work called joule is named after him. This is why we use joule as the unit of both work and energy.
### 10.2.1 Forms of Energy
Luckily, the world we live in provides energy in many different forms. The various forms include mechanical energy, which is the sum of potential energy and kinetic energy, heat energy, chemical energy, electrical energy, and light energy. There are many other forms as well, like sound energy, nuclear energy, and so on. How do you know that some entity is a form of energy? Think about it. Anything that can do work or has the capability to do work is a form of energy. Discuss this with your friends and teachers.
### 10.2.2 Kinetic Energy
Now, students, let us learn about the first type of mechanical energy — kinetic energy. Let me ask you a question. Can a moving object do work? Yes, definitely! A moving object can do work. An object moving faster can do more work than an identical object moving relatively slow. A moving bullet can pierce a target. Wind can move the blades of a windmill. Objects in motion possess energy. We call this energy kinetic energy.
Can you give me some examples of objects that possess kinetic energy? A falling coconut, a speeding car, a rolling stone, a flying aircraft, flowing water, blowing wind, a running athlete — all these possess kinetic energy. In short, kinetic energy is the energy possessed by an object due to its motion. The kinetic energy of an object increases with its speed.
Now, let us do an activity to understand kinetic energy better. Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be better. Drop the ball on the sand bed from a height of about 25 cm. The ball creates a depression. Repeat this activity from heights of 50 cm, 1 m, and 1.5 m. Ensure that all the depressions are distinctly visible. Mark the depressions to indicate the height from which the ball was dropped. Compare their depths. Which one of them is deepest? The one from the greatest height. Which one is shallowest? The one from the smallest height. Why? Because the ball from a greater height has more kinetic energy when it hits the sand, and therefore it can do more work, creating a deeper depression. What has caused the ball to make a deeper dent? The greater kinetic energy of the ball. Discuss and analyse this with your friends.
Now, let us do another activity. Set up the apparatus as shown in Fig. 10.5 in your textbook. Place a wooden block of known mass in front of the trolley at a convenient fixed distance. Place a known mass on the pan so that the trolley starts moving. The trolley moves forward and hits the wooden block. Fix a stop on the table in such a manner that the trolley stops after hitting the block. The block gets displaced. Note down the displacement of the block. This means work is done on the block by the trolley as the block has gained energy. From where does this energy come? It comes from the moving trolley. Repeat this activity by increasing the mass on the pan. In which case is the displacement more? When the mass on the pan is more, the trolley moves faster and hits the block with more force, causing more displacement. In which case is the work done more? In the case where the displacement is more, the work done is more. In this activity, the moving trolley does work and hence it possesses energy. This energy is kinetic energy.
Now, students, how much energy is possessed by a moving body by virtue of its motion? By definition, we say that the kinetic energy of a body moving with a certain velocity is equal to the work done on it to make it acquire that velocity.
Let us now express the kinetic energy of an object in the form of an equation. Consider an object of mass m moving with a uniform velocity u. Let it now be displaced through a distance s when a constant force F acts on it in the direction of its displacement. From equation 10.1, the work done W is F times s. The work done on the object will cause a change in its velocity. Let its velocity change from u to v. Let a be the acceleration produced.
We studied three equations of motion earlier. The relation connecting the initial velocity u and final velocity v of an object moving with a uniform acceleration a, and the displacement s is v squared minus u squared equals 2 a s. This gives s equals v squared minus u squared divided by 2a. This is equation 10.2.
From what we learned in the previous chapter, we know F equals m a. Thus, using equation 10.2 in equation 10.1, we can write the work done by the force F as W equals m a times v squared minus u squared divided by 2a, which simplifies to W equals half m times v squared minus u squared. This is equation 10.3.
If the object is starting from its stationary position, that is, u equals 0, then W equals half m v squared. It is clear that the work done is equal to the change in the kinetic energy of an object. If u equals 0, the work done will be half m v squared. Thus, the kinetic energy possessed by an object of mass m and moving with a uniform velocity v is E subscript k equals half m v squared. This is equation 10.5.
So, students, the kinetic energy of a moving object is given by the formula E subscript k equals half m v squared. This is very important. Remember, kinetic energy depends on the mass of the object and the square of its velocity. This means if you double the velocity, the kinetic energy becomes four times. If you triple the velocity, the kinetic energy becomes nine times. This is why it is so dangerous to drive at high speeds.
Now, let us look at Example 10.3. An object of mass 15 kg is moving with a uniform velocity of 4 m per second. What is the kinetic energy possessed by the object? Mass of the object m equals 15 kg, velocity of the object v equals 4 m/s. From equation 10.5, kinetic energy E subscript k equals half m v squared equals half times 15 kg times 4 m/s times 4 m/s equals 120 J. The kinetic energy of the object is 120 joule.
Now, Example 10.4. What is the work to be done to increase the velocity of a car from 30 km per hour to 60 km per hour if the mass of the car is 1500 kg? Mass of the car m equals 1500 kg. Initial velocity of car u equals 30 km/h. We need to convert this to m/s. 30 km/h equals 30 times 1000 m divided by 60 times 60 s, which equals 25/3 m/s. Similarly, the final velocity of the car v equals 60 km/h equals 50/3 m/s. Therefore, the initial kinetic energy of the car E subscript k i equals half m u squared equals half times 1500 kg times (25/3 m/s) squared equals 156250/3 J. The final kinetic energy of the car E subscript k f equals half times 1500 kg times (50/3 m/s) squared equals 625000/3 J. Thus, the work done equals change in kinetic energy, which is E subscript k f minus E subscript k i, which equals 156250 J. So, the work to be done is 156250 joule.
Now, students, let me ask you the questions from section 10.2.2.
Question 1: What is the kinetic energy of an object? Kinetic energy is the energy possessed by an object due to its motion.
Question 2: Write an expression for the kinetic energy of an object. The expression is E subscript k equals half m v squared.
Question 3: The kinetic energy of an object of mass m moving with a velocity of 5 m/s is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times? First, we need to find the mass of the object. Given that kinetic energy is 25 J when velocity is 5 m/s, we have 25 J equals half m times 5 squared, which means 25 equals half m times 25, so m equals 2 kg. When the velocity is doubled, that is, 10 m/s, kinetic energy equals half times 2 times 10 squared equals 100 J. When the velocity is increased three times, that is, 15 m/s, kinetic energy equals half times 2 times 15 squared equals 225 J.
### 10.2.3 Potential Energy
Now, students, let us learn about the other type of mechanical energy — potential energy. Let me ask you a question. Have you ever stretched a rubber band? What happens when you stretch it and then release it? It snaps back, doesn't it? This means the stretched rubber band has acquired some energy. Let us do some activities to understand potential energy better.
Activity 10.8: Take a rubber band. Hold it at one end and pull from the other. The band stretches. Release the band at one of the ends. What happens? The band will tend to regain its original length. Obviously, the band had acquired energy in its stretched position. How did it acquire energy when stretched? You did work on the rubber band by stretching it, and this work got stored as energy in the stretched rubber band.
Activity 10.9: Take a slinky. Ask a friend to hold one of its ends. You hold the other end and move away from your friend. Now you release the slinky. What happened? The slinky contracted back. How did the slinky acquire energy when stretched? Would the slinky acquire energy when it is compressed? Yes, when you compress a slinky, you also do work on it, and this work gets stored as energy.
Activity 10.10: Take a toy car. Wind it using its key. Place the car on the ground. Did it move? Yes, it moved. From where did it acquire energy? It acquired energy from the winding of the key. Does the energy acquired depend on the number of windings? Yes, more windings mean more energy stored. How can you test this? Try winding the key different numbers of times and see how far the car goes.
Activity 10.11: Lift an object through a certain height. The object can now do work. It begins to fall when released. This implies that it has acquired some energy. If raised to a greater height, it can do more work and hence possesses more energy. From where did it get the energy? Think and discuss. The energy came from the work done in lifting the object against gravity.
In the above situations, the energy gets stored due to the work done on the object. The energy transferred to an object is stored as potential energy if it is not used to cause a change in the velocity or speed of the object.
You transfer energy when you stretch a rubber band. The energy transferred to the band is its potential energy. You do work while winding the key of a toy car. The energy transferred to the spring inside is stored as potential energy. The potential energy possessed by the object is the energy present in it by virtue of its position or configuration.
Let me give you another example. Activity 10.12: Take a bamboo stick and make a bow. Place an arrow made of a light stick on it with one end supported by the stretched string. Now stretch the string and release the arrow. Notice the arrow flying off the bow. Notice the change in the shape of the bow. The potential energy stored in the bow due to the change of shape is thus used in the form of kinetic energy in throwing off the arrow. This is exactly how an arrow is shot from a bow!
### 10.2.4 Potential Energy of an Object at a Height
Now, students, let us specifically talk about gravitational potential energy. An object increases its energy when raised through a height. This is because work is done on it against gravity while it is being raised. The energy present in such an object is the gravitational potential energy.
The gravitational potential energy of an object at a point above the ground is defined as the work done in raising it from the ground to that point against gravity.
It is easy to arrive at an expression for the gravitational potential energy of an object at a height. Consider an object of mass m. Let it be raised through a height h from the ground. A force is required to do this. The minimum force required to raise the object is equal to the weight of the object, mg. The object gains energy equal to the work done on it. Let the work done on the object against gravity be W. That is, work done W equals force times displacement equals mg times h equals mgh. Since work done on the object is equal to mgh, an energy equal to mgh units is gained by the object. This is the potential energy E subscript p of the object. So, E subscript p equals mgh. This is equation 10.6.
Now, students, there is an important point to note. The potential energy of an object at a height depends on the ground level or the zero level you choose. An object in a given position can have a certain potential energy with respect to one level and a different value of potential energy with respect to another level. For example, if you are standing on the first floor of a building, your potential energy with respect to the ground floor is different from your potential energy with respect to the second floor. So, it is important to define the reference level or the zero level for potential energy.
It is also useful to note that the work done by gravity depends on the difference in vertical heights of the initial and final positions of the object and not on the path along which the object is moved. Your textbook shows a case where a block is raised from position A to B by taking two different paths. Let the height AB equal h. In both the situations, the work done on the object is mgh. This is because gravity is a conservative force, and the work done by gravity depends only on the initial and final positions, not on the path taken.
Now, let us look at Example 10.5. Find the energy possessed by an object of mass 10 kg when it is at a height of 6 m above the ground. Given, g equals 9.8 m/s². Mass of the object m equals 10 kg, displacement (height) h equals 6 m, and acceleration due to gravity g equals 9.8 m/s². From equation 10.6, potential energy equals mgh equals 10 kg times 9.8 m/s² times 6 m equals 588 J. The potential energy is 588 joule.
Example 10.6: An object of mass 12 kg is at a certain height above the ground. If the potential energy of the object is 480 J, find the height at which the object is with respect to the ground. Given, g equals 10 m/s². Mass of the object m equals 12 kg, potential energy E subscript p equals 480 J. We have E subscript p equals mgh, so 480 J equals 12 kg times 10 m/s² times h. Therefore, h equals 480 J divided by 120 kg m/s² equals 4 m. The object is at the height of 4 m.
### 10.2.5 Are Various Energy Forms Interconvertible?
Now, students, can we convert energy from one form to another? We find in nature a number of instances of conversion of energy from one form to another. Let us do an activity to understand this better.
Activity 10.13: Sit in small groups. Discuss the various ways of energy conversion in nature. Discuss the following questions in your group: (a) How do green plants produce food? They use sunlight to convert carbon dioxide and water into glucose and oxygen. This is photosynthesis, where light energy is converted into chemical energy. (b) Where do they get their energy from? From the Sun. (c) Why does the air move from place to place? Because of differences in temperature and pressure, which are caused by solar energy. (d) How are fuels such as coal and petroleum formed? They are formed from the remains of ancient plants and animals that lived millions of years ago. These organisms got their energy from the Sun, and over time, this solar energy got stored in fossil fuels. (e) What kinds of energy conversions sustain the water cycle? The Sun evaporates water, which then condenses and falls as rain. This involves conversion of solar energy into potential energy of water vapor, which then converts to kinetic energy as rain falls.
Activity 10.14: Many of the human activities and the gadgets we use involve conversion of energy from one form to another. Make a list of such activities and gadgets. Identify in each activity or gadget the kind of energy conversion that takes place. For example, a fan converts electrical energy into mechanical energy (kinetic energy). A light bulb converts electrical energy into light energy. A stove converts chemical energy (from LPG) into heat energy. A microphone converts sound energy into electrical energy. A solar panel converts light energy into electrical energy. There are countless examples.
### 10.2.6 Law of Conservation of Energy
Now, students, this is a very important concept. In activities 10.13 and 10.14, we learnt that the form of energy can be changed from one form to another. What happens to the total energy of a system during or after the process? Whenever energy gets transformed, the total energy remains unchanged. This is the law of conservation of energy. According to this law, energy can only be converted from one form to another; it can neither be created nor destroyed. The total energy before and after the transformation always remains the same. The law of conservation of energy is valid in all situations and for all kinds of transformations.
Consider a simple example. Let an object of mass m be made to fall freely from a height h. At the start, the potential energy is mgh and kinetic energy is zero. Why is the kinetic energy zero? It is zero because its velocity is zero. The total energy of the object is thus mgh. As it falls, its potential energy will change into kinetic energy. If v is the velocity of the object at a given instant, the kinetic energy would be half m v squared. As the fall of the object continues, the potential energy would decrease while the kinetic energy would increase. When the object is about to reach the ground, h equals 0 and v will be the highest. Therefore, the kinetic energy would be the largest and potential energy the least. However, the sum of the potential energy and kinetic energy of the object would be the same at all points. That is, potential energy plus kinetic energy equals constant, or mgh plus half m v squared equals constant. This is equation 10.7.
The sum of kinetic energy and potential energy of an object is its total mechanical energy. We find that during the free fall of the object, the decrease in potential energy at any point in its path appears as an equal amount of increase in kinetic energy. Here the effect of air resistance on the motion of the object has been ignored. There is thus a continual transformation of gravitational potential energy into kinetic energy.
Now, let us do Activity 10.15. An object of mass 20 kg is dropped from a height of 4 m. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case. For simplifying the calculations, take the value of g as 10 m/s².
At height 4 m: Potential energy E subscript p equals mgh equals 20 kg times 10 m/s² times 4 m equals 800 J. Kinetic energy is 0 because the object is just about to be dropped. Total energy E subscript p plus E subscript k equals 800 J.
At height 3 m: Potential energy equals 20 times 10 times 3 equals 600 J. To find kinetic energy, we need to find the velocity at height 3 m. We can use the equation v squared equals u squared plus 2gh, where u is initial velocity (0), g is 10, and h is the distance fallen (4 minus 3 equals 1 m). So v squared equals 0 plus 2 times 10 times 1 equals 20, so v equals square root of 20 which is about 4.47 m/s. Kinetic energy equals half times 20 times 20 equals 200 J. Total energy equals 600 plus 200 equals 800 J.
At height 2 m: Potential energy equals 20 times 10 times 2 equals 400 J. Distance fallen is 2 m, so v squared equals 2 times 10 times 2 equals 40, v equals square root of 40 which is about 6.32 m/s. Kinetic energy equals half times 20 times 40 equals 400 J. Total energy equals 400 plus 400 equals 800 J.
At height 1 m: Potential energy equals 20 times 10 times 1 equals 200 J. Distance fallen is 3 m, so v squared equals 2 times 10 times 3 equals 60, v equals square root of 60 which is about 7.75 m/s. Kinetic energy equals half times 20 times 60 equals 600 J. Total energy equals 200 plus 600 equals 800 J.
Just above the ground: Potential energy equals 0 (since h equals 0). Distance fallen is 4 m, so v squared equals 2 times 10 times 4 equals 80, v equals square root of 80 which is about 8.94 m/s. Kinetic energy equals half times 20 times 80 equals 800 J. Total energy equals 0 plus 800 equals 800 J.
So, students, you can see that the total energy remains constant at all points. This is the law of conservation of energy in action.
Now, students, think about this. What would have happened if nature had not allowed the transformation of energy? There is a view that life could not have been possible without transformation of energy. Do you agree with this? Think about it and discuss with your friends and teacher.
## 10.3 Rate of Doing Work
Now, students, let us learn about power, which is the rate of doing work. Do all of us work at the same rate? Do machines consume or transfer energy at the same rate? Agents that transfer energy do work at different rates. Let us understand this from the following activity.
Activity 10.16: Consider two children, say A and B. Let us say they weigh the same. Both start climbing up a rope separately. Both reach a height of 8 m. Let us say A takes 15 s while B takes 20 s to accomplish the task. What is the work done by each? The work done is the same because they both lift the same weight through the same height. However, A has taken less time than B to do the work. Who has done more work in a given time, say in 1 s? A has done more work in 1 s because A completed the same work in less time. So, A has more power than B. Power measures how fast work is done.
Power is defined as the rate of doing work or the rate of transfer of energy. If an agent does a work W in time t, then power is given by Power equals work divided by time, or P equals W divided by t. This is equation 10.8.
The unit of power is watt, in honour of James Watt, having the symbol W. 1 watt is the power of an agent which does work at the rate of 1 joule per second. We can also say that power is 1 W when the rate of consumption of energy is 1 J/s.
1 watt equals 1 joule/second, or 1 W equals 1 J/s.
We express larger rates of energy transfer in kilowatts. 1 kilowatt equals 1000 watts, 1 kW equals 1000 W, and 1 kW equals 1000 J/s.
The power of an agent may vary with time. This means that the agent may be doing work at different rates at different intervals of time. Therefore, the concept of average power is useful. We obtain average power by dividing the total energy consumed by the total time taken.
Now, let us look at Example 10.7. Two girls, each of weight 400 N climb up a rope through a height of 8 m. We name one of the girls A and the other B. Girl A takes 20 s while B takes 50 s to accomplish this task. What is the power expended by each girl?
Solution for girl A: Weight of the girl mg equals 400 N, displacement (height) h equals 8 m, time taken t equals 20 s. From equation 10.8, power P equals work done divided by time taken equals mgh divided by t equals 400 N times 8 m divided by 20 s equals 160 W. Power expended by girl A is 160 W.
Solution for girl B: Weight of the girl mg equals 400 N, displacement (height) h equals 8 m, time taken t equals 50 s. Power P equals mgh divided by t equals 400 N times 8 m divided by 50 s equals 64 W. Power expended by girl B is 64 W.
Example 10.8: A boy of mass 50 kg runs up a staircase of 45 steps in 9 s. If the height of each step is 15 cm, find his power. Take g equals 10 m/s².
Solution: Weight of the boy mg equals 50 kg times 10 m/s² equals 500 N. Height of the staircase h equals 45 times 15/100 m equals 6.75 m. Time taken to climb t equals 9 s. From equation 10.8, power P equals work done divided by time taken equals mgh divided by t equals 500 N times 6.75 m divided by 9 s equals 375 W. Power is 375 W.
Now, students, let me ask you the questions from section 10.3.
Question 1: What is power? Power is defined as the rate of doing work or the rate of transfer of energy.
Question 2: Define 1 watt of power. 1 watt is the power of an agent which does work at the rate of 1 joule per second.
Question 3: A lamp consumes 1000 J of electrical energy in 10 s. What is its power? Power equals work divided by time equals 1000 J divided by 10 s equals 100 W.
Question 4: Define average power. Average power is obtained by dividing the total energy consumed by the total time taken.
Now, students, there is an interesting activity you can do at home. Activity 10.17: Take a close look at the electric meter installed in your house. Observe its features closely. Take the readings of the meter each day at 6:30 am and 6:30 pm. Do this activity for about a week. How many units are consumed during daytime? How many units are used during night? Tabulate your observations. Draw inferences from the data. Compare your observations with the details given in the monthly electricity bill. One can also estimate the electricity to be consumed by specific appliances by tabulating their known wattages and hours of operation. This will help you understand how power and energy are related in real life.
Now, students, let us solve the exercises at the end of the chapter. These are very important questions, and you should practice all of them.
Exercise 1: Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term work.
(a) Suma is swimming in a pond. Yes, work is done. Suma is applying force on the water with her arms and legs, and she is moving through the water. So, there is force and displacement.
(b) A donkey is carrying a load on its back. Yes, work is done. The donkey is applying an upward force to support the load, and the donkey is moving forward. So, there is force and displacement.
(c) A wind-mill is lifting water from a well. Yes, work is done. The wind is applying force on the blades of the windmill, causing it to rotate and lift water. So, there is force and displacement.
(d) A green plant is carrying out photosynthesis. This is a biochemical process. In the scientific sense, no mechanical work is done because there is no macroscopic force and displacement involved. However, from a biological perspective, energy transformation is occurring.
(e) An engine is pulling a train. Yes, work is done. The engine applies force on the train, and the train moves. So, there is force and displacement.
(f) Food grains are getting dried in the sun. No work is done in the scientific sense. The sun's energy is being transferred to the grains, but there is no force applied and no displacement of any object due to an applied force.
(g) A sailboat is moving due to wind energy. Yes, work is done. The wind applies force on the sail, and the boat moves. So, there is force and displacement.
Exercise 2: An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object? The work done by gravity is zero because the initial and final vertical heights are the same. Work done by gravity equals mg times change in height, and since the change in height is zero, work done is zero.
Exercise 3: A battery lights a bulb. Describe the energy changes involved in the process. The battery has chemical energy. When the battery is connected to the bulb, the chemical energy is converted into electrical energy. This electrical energy is then converted into light energy and heat energy by the bulb.
Exercise 4: Certain force acting on a 20 kg mass changes its velocity from 5 m/s to 2 m/s. Calculate the work done by the force. Initial velocity u equals 5 m/s, final velocity v equals 2 m/s, mass m equals 20 kg. Initial kinetic energy equals half m u squared equals half times 20 times 25 equals 250 J. Final kinetic energy equals half m v squared equals half times 20 times 4 equals 40 J. Work done equals change in kinetic energy equals final kinetic energy minus initial kinetic energy equals 40 minus 250 equals minus 210 J. The work done is negative because the force is acting opposite to the direction of motion, causing the object to slow down.
Exercise 5: A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer. The work done is zero because the gravitational force acts vertically downwards, while the displacement is horizontal. There is no component of gravitational force in the direction of displacement. So, work done by gravity is zero.
Exercise 6: The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why? No, it does not violate the law of conservation of energy. The decrease in potential energy is exactly equal to the increase in kinetic energy. The total mechanical energy (potential energy plus kinetic energy) remains constant. So, energy is conserved.
Exercise 7: What are the various energy transformations that occur when you are riding a bicycle? When you ride a bicycle, you use the chemical energy from the food you ate. This chemical energy is converted into mechanical energy as your legs push the pedals. This mechanical energy is transferred to the bicycle, which is then converted into kinetic energy of the bicycle and the rider. Some energy is also converted into heat energy due to friction.
Exercise 8: Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going? Yes, energy is being transferred, but no work is done in the scientific sense because there is no displacement. The energy you spend is converted into heat energy in your muscles and body. You feel tired because your muscles are converting chemical energy into heat energy.
Exercise 9: A certain household has consumed 250 units of energy during a month. How much energy is this in joules? In electricity billing, 1 unit means 1 kilowatt-hour. 1 kilowatt-hour equals 1000 watts times 3600 seconds equals 3.6 times 10 to the power of 6 joules. So, 250 units equal 250 times 3.6 times 10 to the power of 6 joules equals 9 times 10 to the power of 8 joule.
Exercise 10: An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down. Potential energy at height 5 m equals mgh equals 40 kg times 10 m/s² times 5 m equals 2000 J. When the object is halfway down, the height is 2.5 m. At this point, potential energy equals 40 times 10 times 2.5 equals 1000 J. Since total energy is conserved and equals 2000 J, kinetic energy at halfway point equals total energy minus potential energy equals 2000 minus 1000 equals 1000 J.
Exercise 11: What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer. The work done by gravity on a satellite moving in a circular orbit is zero. This is because the force of gravity is always perpendicular to the displacement of the satellite. Since work done equals force times displacement times cos theta, and theta equals 90 degrees, cos 90 equals 0, so work done is zero.
Exercise 12: Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher. Yes, according to Newton's first law of motion, an object will continue to be in a state of rest or uniform motion in a straight line in the absence of any external force. So, if an object is already moving, it can continue to move with constant velocity even without any force acting on it. However, in most real situations, there are forces like friction and air resistance that slow down moving objects.
Exercise 13: A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer. In the scientific sense, no work is done because there is no displacement of the bundle of hay. The person is exerting an upward force to hold the bundle, but the bundle is not moving. So, work done is zero. However, the person feels tired because his muscles are doing internal work to maintain the tension, and this internal work produces heat energy.
Exercise 14: An electric heater is rated 1500 W. How much energy does it use in 10 hours? Power P equals 1500 W equals 1500 J/s. Time t equals 10 hours equals 10 times 3600 seconds equals 36000 s. Energy consumed equals P times t equals 1500 J/s times 36000 s equals 54 times 10 to the power of 6 J, which is 54 MJ. Alternatively, in kilowatt-hours, energy equals 1.5 kW times 10 h equals 15 kWh, which is 15 units.
Exercise 15: Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy? When we draw the pendulum bob to one side, we do work on it against gravity. This work gets stored as gravitational potential energy. When we release the bob, it starts moving towards the mean position. As it moves, potential energy decreases and kinetic energy increases. At the mean position, potential energy is minimum and kinetic energy is maximum. The bob overshoots the mean position and moves to the other side. As it moves up, kinetic energy decreases and potential energy increases. At the extreme positions, kinetic energy is zero and potential energy is maximum. This process continues, and energy transforms back and forth between potential and kinetic forms. However, the bob eventually comes to rest because of air resistance and friction at the pivot. The energy is not lost but is converted into heat energy in the air and the pivot. So, the total energy is still conserved; it is just converted from mechanical energy to heat energy. This is not a violation of the law of conservation of energy.
Exercise 16: An object of mass m is moving with a constant velocity v. How much work should be done on the object in order to bring the object to rest? The kinetic energy of the moving object is half m v squared. To bring the object to rest, we need to remove all this kinetic energy. So, the work done should be equal to half m v squared, but in the negative direction. So, the work done should be minus half m v squared.
Exercise 17: Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h. First, convert 60 km/h to m/s. 60 km/h equals 60 times 1000 divided by 3600 equals 16.67 m/s approximately. Kinetic energy of the car equals half times 1500 times (16.67)² equals about 208333 J. To stop the car, we need to do this much work against its motion. So, the work required is about 208333 J.
Exercise 18: In each of the following, a force F is acting on an object of mass m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive, or zero. Since I cannot see the diagrams, I will describe what to look for. If the force is in the same direction as displacement (west to east), work is positive. If the force is opposite to displacement (east to west), work is negative. If the force is perpendicular to displacement (north or south), work is zero.
Exercise 19: Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why? Yes, I agree with Soni. Acceleration is zero when the net force acting on an object is zero. Several forces can act on an object, but if they cancel each other out, the net force is zero, and acceleration is zero. For example, if you push a box with a force of 10 N to the right, and your friend pushes it with a force of 10 N to the left, the net force is zero, and the box does not accelerate.
Exercise 20: Find the energy in joules consumed in 10 hours by four devices of power 500 W each. Power of each device equals 500 W. Power of four devices equals 4 times 500 W equals 2000 W. Time equals 10 hours equals 10 times 3600 seconds equals 36000 s. Energy consumed equals power times time equals 2000 W times 36000 s equals 72 times 10^6 J, which is 72 MJ.
Exercise 21: A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy? When a freely falling object hits the ground, its kinetic energy is transferred to the ground or converted into other forms of energy. If the object bounces, some kinetic energy is converted into potential energy and sound energy. If the object does not bounce, the kinetic energy is converted into heat energy and sound energy. In all cases, energy is conserved; it is just transformed from one form to another.
Now, students, we have covered the entire chapter. Let me give you a quick summary of everything we have learned.
## Summary
In this chapter, we learned about work, energy, and power.
Work is done when a force acts on an object and the object is displaced in the direction of the force. Work is calculated as the product of force and displacement: W equals F times s. The unit of work is joule. Work can be positive or negative depending on whether the force is in the same direction as displacement or opposite to it.
Energy is the capability of an object to do work. The unit of energy is also joule. There are many forms of energy, including mechanical energy (kinetic energy plus potential energy), heat energy, chemical energy, electrical energy, and light energy.
Kinetic energy is the energy possessed by an object due to its motion. The kinetic energy of an object of mass m moving with velocity v is given by E subscript k equals half m v squared.
Potential energy is the energy possessed by an object due to its position or configuration. Gravitational potential energy is given by E subscript p equals m g h, where h is the height above the ground.
According to the law of conservation of energy, energy can only be transformed from one form to another; it can neither be created nor destroyed. The total energy before and after the transformation always remains the same.
Power is the rate of doing work or the rate of transfer of energy. Power equals work divided by time, or P equals W divided by t. The unit of power is watt. 1 W equals 1 J/s.
Students, I hope this lesson has helped you understand the concepts of work and energy thoroughly. Remember, these concepts are very important not only for your exams but also for understanding the world around you. Energy is everywhere, and understanding how it works helps us appreciate the wonders of nature and technology. Keep learning, keep exploring, and never stop asking questions. Thank you for your attention, and have a great day!