Welcome dear students! Today we are going to learn about Work and Energy from Class 9 Science. In the previous few chapters we have talked about ways of describing the motion of objects, the cause of motion and gravitation. Another concept that helps us understand and interpret many natural phenomena is work. Closely related to work are energy and power. In this chapter we shall study these concepts. All living beings need food. Living beings have to perform several basic activities to survive. We call such activities life processes. The energy for these processes comes from food. We need energy for other activities like playing, singing, reading, writing, thinking, jumping, cycling and running. Activities that are strenuous require more energy. Animals too get engaged in activities. For example, they may jump and run. They have to fight, move away from enemies, find food or find a safe place to live. Also, we engage some animals to lift weights, carry loads, pull carts or plough fields. All such activities require energy. Think of machines. List the machines that you have come across. What do they need for their working? Why do some engines require fuel like petrol and diesel? Why do living beings and machines need energy? [CHECKPOINT]
Let us begin with section ten point one, Work. What is work? There is a difference in the way we use the term work in day-to-day life and the way we use it in science. To make this point clear let us consider a few examples. Kamali is preparing for examinations. She spends lot of time in studies. 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, she is working hard. All this hard work may involve very little work if we go by the scientific definition of work. You are working hard to push a huge rock. Let us say the rock does not move despite all the effort. You get completely exhausted. However, you have not done any work on the rock as there is no displacement of the rock. You stand still for a few minutes with a heavy load on your head. You get tired. You have exerted yourself and have spent quite a bit of your energy. Are you doing work on the load? The way we understand the term work in science, work is not done. You climb up the steps of a staircase and reach the second floor of a building just to see the landscape from there. You may even climb up a tall tree. If we apply the scientific definition, these activities involve a lot of work. In day-to-day life, we consider any useful physical or mental labour as work. Activities like playing in a field, talking with friends, humming a tune, watching a movie, attending a function are sometimes not considered to be work. What constitutes work depends on the way we define it. We use and define the term work differently in science. [CHECKPOINT]
To understand this let us do Activity ten point one. We have discussed a number of activities which we normally consider to be work in day-to-day life. For each of these activities, ask the following questions and answer them. What is the work being done on? What is happening to the object? Who or what is doing the work? Now let us move to the scientific conception of work. To understand the way we view work and define work from the point of view of science, let us consider some situations. 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. A girl pulls a trolley and the trolley moves through a distance. The girl has exerted a force on the trolley and it is displaced. Therefore, work is done. Lift 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. A closer look at the above situations reveals that two conditions need to be satisfied for work to be done. First, a force should act on an object, and second, the object must be displaced. If any one of the above conditions does not exist, work is not done. This is the way we view work in science. 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? [CHECKPOINT]
Activity ten point two asks you to think of some situations from your daily life involving work, list them, and discuss with your friends whether work is being done in each situation. Try to reason out your response. If work is done, identify the force acting on the object, the object on which the work is done, and what happens to that object. Activity ten point three asks you to think of situations when the object is not displaced in spite of a force acting on it, and situations when an object gets displaced in the absence of a force acting on it. List all situations and discuss whether work is done. Now, how is work defined in science? To understand this, we shall first consider the case when the force is acting in the direction of displacement. Let a constant force, F act on an object. Let the object be displaced through a distance, s in the direction of the force. In Figure ten point one, we see an arrow labeled F pointing to the right, and an object moving a distance s in the same direction. Let W be the work done. We define work to be equal to the product of the force and displacement. Work done equals force multiplied by displacement. In symbols, W equals F s. 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. In this equation, if F equals one newton and s equals one metre then the work done by the force will be one newton metre. Here the unit of work is newton metre or joule. Thus one joule is the amount of work done on an object when a force of one newton displaces it by one metre along the line of action of the force. [CHECKPOINT]
Look at the equation carefully. What is the work done when the force on the object is zero? What would be the work done when the displacement of the object is zero? Refer to the conditions that are to be satisfied to say that work is done. Let us solve Example ten point one. A force of five newtons is acting on an object. The object is displaced through two metres in the direction of the force. In Figure ten point two, we see a force vector of five newtons pointing right, and a displacement of two metres in the same direction. If the force acts on the object all through the displacement, then work done is five newtons multiplied by two metres, which equals ten newton metres or ten joules. Question one asks: A force of seven newtons acts on an object. The displacement is eight metres in the direction of the force. In Figure ten point three, we see a seven newton force and an eight metre displacement in the same direction. Let us take it that the force acts on the object through the displacement. What is the work done in this case? The work done is seven newtons multiplied by eight metres, which equals fifty-six joules. Consider another situation where force and displacement are in the same direction: a baby pulling a toy car parallel to the ground, as shown in Figure ten point four. The diagram shows a baby pulling a string attached to a car, with force and displacement arrows pointing right. The baby has exerted a force in the direction of displacement of the car. In this situation, the work done will be equal to the product of the force and displacement. In such situations, the work done by the force is taken as positive. [CHECKPOINT]
Consider a situation in which an object is moving with a uniform velocity along a particular direction. Now a retarding force, F, is applied in the opposite direction. That is, the angle between the two directions is one hundred eighty degrees. Let the object stop after a displacement s. In such a situation, the work done by the force, F is taken as negative and denoted by the minus sign. The work done by the force is F multiplied by minus s, or minus F multiplied by s. It is clear that the work done by a force can be either positive or negative. To understand this, let us do Activity ten point four. 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 the force of gravity acting on the object. Which one of these forces is doing positive work? Which one is doing negative work? Give reasons. Work done is negative when the force acts opposite to the direction of displacement. Work done is positive when the force is in the direction of displacement. Let us solve Example ten point two. A porter lifts a luggage of fifteen kilograms from the ground and puts it on his head one point five metres above the ground. Calculate the work done by him on the luggage. Mass of luggage, m equals fifteen kilograms and displacement, s equals one point five metres. Work done, W equals F multiplied by s, which is mg multiplied by s. This equals fifteen kilograms multiplied by ten metres per second squared multiplied by one point five metres. This gives two hundred twenty-five kilogram metres per second squared metres, which is two hundred twenty-five newton metres, or two hundred twenty-five joules. Work done is two hundred twenty-five joules. [CHECKPOINT]
Questions for this section: One, when do we say that work is done? We say work is done when a force acts on an object and the object is displaced in the direction of the force. Two, write an expression for the work done when a force is acting on an object in the direction of its displacement. The expression is W equals F s. Three, define one joule of work. One joule is the amount of work done on an object when a force of one newton displaces it by one metre along the line of action of the force. Four, a pair of bullocks exerts a force of one hundred forty newtons on a plough. The field being ploughed is fifteen metres long. How much work is done in ploughing the length of the field? Work done equals force multiplied by displacement, which is one hundred forty newtons multiplied by fifteen metres, giving two thousand one hundred joules. [CHECKPOINT]
Let us move to section ten point two, 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. We can also get energy from the nuclei of atoms, the interior of the earth, and the tides. Can you think of other sources of energy? Activity ten point five asks you to list other sources of energy, discuss in small groups how certain sources are due to the Sun, and identify sources not due to the Sun. 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 examples. When a fast moving cricket ball hits a stationary wicket, the wicket is thrown away. 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. 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. 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. One joule is the energy required to do one joule of work. Sometimes a larger unit of energy called kilo joule is used. One kilo joule equals one thousand joules. [CHECKPOINT]
Section ten point two point one covers Forms of Energy. Luckily the world we live in provides energy in many different forms. The various forms include mechanical energy, which is potential energy plus kinetic energy, heat energy, chemical energy, electrical energy and light energy. Think it over: How do you know that some entity is a form of energy? Discuss with your friends and teachers. Did you know that James Prescott Joule was an outstanding British physicist? He lived from eighteen eighteen to eighteen eighty-nine. 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. Let us look at Activity ten point six. 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 height of about twenty-five centimetres. The ball creates a depression. Repeat this activity from heights of fifty centimetres, one metre and one point five metres. 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? Which one is shallowest? Why? What has caused the ball to make a deeper dent? Discuss and analyse. The ball dropped from the greatest height makes the deepest dent because it possesses more energy due to its greater height. [CHECKPOINT]
Now, section ten point two point two, Kinetic Energy. Activity ten point seven asks you to set up the apparatus as shown in Figure ten point five. In this diagram, we see a trolley on a horizontal table. A string is attached to the trolley, passes over a pulley at the edge of the table, and has a pan hanging vertically. A wooden block is placed in front of the trolley. A stop is fixed on the table. 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? Repeat this activity by increasing the mass on the pan. In which case is the displacement more? In which case is the work done more? In this activity, the moving trolley does work and hence it possesses energy. A moving object can do work. An object moving faster can do more work than an identical object moving relatively slow. A moving bullet, blowing wind, a rotating wheel, a speeding stone can do work. How does a bullet pierce the target? How does the wind move the blades of a windmill? Objects in motion possess energy. We call this energy kinetic energy. A falling coconut, a speeding car, a rolling stone, a flying aircraft, flowing water, blowing wind, a running athlete etc. 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. 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. [CHECKPOINT]
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 the work equation, the work done, W is F 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. 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 two a s. This gives s equals v squared minus u squared divided by two a. From section nine point four, we know F equals m a. Thus, using this in the work equation, we can write the work done by the force, F as W equals m a multiplied by v squared minus u squared divided by two a, or W equals one half m times v squared minus u squared. If the object is starting from its stationary position, that is, u equals zero, then W equals one 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 zero, the work done will be one half m v squared. Thus, the kinetic energy possessed by an object of mass, m and moving with a uniform velocity, v is E sub k equals one half m v squared. [CHECKPOINT]
Let us solve Example ten point three. An object of mass fifteen kilograms is moving with a uniform velocity of four metres per second. What is the kinetic energy possessed by the object? Mass of the object, m equals fifteen kilograms, velocity of the object, v equals four metres per second. From the equation, E sub k equals one half m v squared, which is one half multiplied by fifteen kilograms multiplied by four metres per second multiplied by four metres per second. This equals one hundred twenty joules. The kinetic energy of the object is one hundred twenty joules. Example ten point four asks: What is the work to be done to increase the velocity of a car from thirty kilometres per hour to sixty kilometres per hour if the mass of the car is fifteen hundred kilograms? Mass of the car, m equals fifteen hundred kilograms. Initial velocity of car, u equals thirty kilometres per hour, which converts to twenty-five thirds metres per second. Similarly, the final velocity of the car, v equals sixty kilometres per hour, which is fifty thirds metres per second. Therefore, the initial kinetic energy of the car, E sub ki equals one half m u squared, which is one half multiplied by fifteen hundred kilograms multiplied by twenty-five thirds metres per second squared. This equals one hundred fifty-six thousand two hundred fifty thirds joules. The final kinetic energy of the car, E sub kf equals one half multiplied by fifteen hundred kilograms multiplied by fifty thirds metres per second squared. This equals six hundred twenty-five thousand thirds joules. Thus, the work done equals change in kinetic energy, which is E sub kf minus E sub ki, giving one hundred fifty-six thousand two hundred fifty joules. [CHECKPOINT]
Questions for kinetic energy: One, what is the kinetic energy of an object? It is the energy possessed by an object due to its motion. Two, write an expression for the kinetic energy of an object. The expression is E sub k equals one half m v squared. Three, the kinetic energy of an object of mass, m moving with a velocity of five metres per second is twenty-five joules. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times? Since kinetic energy is proportional to the square of velocity, doubling the velocity makes the kinetic energy four times, so it becomes one hundred joules. Tripling the velocity makes it nine times, so it becomes two hundred twenty-five joules. Now, Activity ten point eight. 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? Activity ten point nine. Take a slinky as shown below. 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. The diagram shows two hands stretching a slinky horizontally. What happened? The slinky contracts back. How did the slinky acquire energy when stretched? Would the slinky acquire energy when it is compressed? Yes, it acquires energy in both stretched and compressed states. [CHECKPOINT]
Activity ten point ten. Take a toy car. Wind it using its key. Place the car on the ground. Did it move? Yes. From where did it acquire energy? From the wound spring inside. Does the energy acquired depend on the number of windings? Yes. How can you test this? By winding it different numbers of times and measuring the distance it travels. Activity ten point eleven. 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. 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. Activity ten point twelve. Take a bamboo stick and make a bow as shown in Figure ten point six. The diagram shows a curved bamboo bow with a stretched string and an arrow resting on it. 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. [CHECKPOINT]
Section ten point two point three is Potential Energy, and ten point two point four covers Potential Energy of an Object at a Height. 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. Figure ten point seven shows a diagram with a block resting on the ground, and an identical block raised to a height h above the ground. A vertical arrow indicates the height h. 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 multiplied by displacement, which is mg multiplied by h, giving 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 sub p, of the object. E sub p equals mgh. [CHECKPOINT]
More to know: 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. It is 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. Figure ten point eight shows a case where a block is raised from position A to B by taking two different paths. The diagram shows Path one as a vertical straight line and Path two as a stepped path, both going from A to B, with height h indicated. In both the situations the work done on the object is mgh. Let us solve Example ten point five. Find the energy possessed by an object of mass ten kilograms when it is at a height of six metres above the ground. Given, g equals nine point eight metres per second squared. Mass of the object, m equals ten kilograms, displacement, h equals six metres, and acceleration due to gravity, g equals nine point eight metres per second squared. From the equation, Potential energy equals mgh, which is ten kilograms multiplied by nine point eight metres per second squared multiplied by six metres. This equals five hundred eighty-eight joules. The potential energy is five hundred eighty-eight joules. [CHECKPOINT]
Example ten point six. An object of mass twelve kilograms is at a certain height above the ground. If the potential energy of the object is four hundred eighty joules, find the height at which the object is with respect to the ground. Given, g equals ten metres per second squared. Mass of the object, m equals twelve kilograms, potential energy, E sub p equals four hundred eighty joules. Using E sub p equals mgh, we get four hundred eighty joules equals twelve kilograms multiplied by ten metres per second squared multiplied by h. Solving for h, we divide four hundred eighty joules by one hundred twenty kilogram metres per second squared, which gives four metres. The object is at the height of four metres. Section ten point two point five asks: Are various energy forms interconvertible? 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. Activity ten point thirteen asks you to sit in small groups, discuss the various ways of energy conversion in nature, and answer questions: How do green plants produce food? Where do they get their energy from? Why does the air move from place to place? How are fuels, such as coal and petroleum formed? What kinds of energy conversions sustain the water cycle? Activity ten point fourteen. 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. [CHECKPOINT]
Section ten point two point six is the Law of Conservation of Energy. In activities ten point thirteen and ten point fourteen, 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 or destroyed. The total energy before and after the transformation 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 one half mv 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 zero 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 one half mv squared equals constant. 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. [CHECKPOINT]
Activity ten point fifteen. An object of mass twenty kilograms is dropped from a height of four metres. 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 ten metres per second squared. At height four metres, potential energy is mgh, which is twenty times ten times four, giving eight hundred joules. Kinetic energy is zero. Sum is eight hundred joules. At height three metres, potential energy is twenty times ten times three, giving six hundred joules. Kinetic energy is two hundred joules. Sum is eight hundred joules. At height two metres, potential energy is four hundred joules. Kinetic energy is four hundred joules. Sum is eight hundred joules. At height one metre, potential energy is two hundred joules. Kinetic energy is six hundred joules. Sum is eight hundred joules. Just above the ground at zero metres, potential energy is zero. Kinetic energy is eight hundred joules. Sum is eight hundred joules. Think it over: 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? [CHECKPOINT]
Let us move to section ten point three, 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 Activity ten point sixteen. 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 eight metres. Let us say A takes fifteen seconds while B takes twenty seconds to accomplish the task. What is the work done by each? The work done is the same. However, A has taken less time than B to do the work. Who has done more work in a given time, say in one second? A has done more work per second. A stronger person may do certain work in relatively less time. A more powerful vehicle would complete a journey in a shorter time than a less powerful one. We talk of the power of machines like motorbikes and motorcars. The speed with which these vehicles change energy or do work is a basis for their classification. Power measures the speed of work done, that is, how fast or slow 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. The unit of power is watt, in honour of James Watt, who lived from seventeen thirty-six to eighteen nineteen, having the symbol W. One watt is the power of an agent, which does work at the rate of one joule per second. We can also say that power is one watt when the rate of consumption of energy is one joule per second. One watt equals one joule per second. We express larger rates of energy transfer in kilowatts. One kilowatt equals one thousand watts. One kilowatt equals one thousand joules per second. 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. [CHECKPOINT]
Let us solve Example ten point seven. Two girls, each of weight four hundred newtons climb up a rope through a height of eight metres. We name one of the girls A and the other B. Girl A takes twenty seconds while B takes fifty seconds to accomplish this task. What is the power expended by each girl? For girl A: Weight of the girl, mg equals four hundred newtons. Displacement, h equals eight metres. Time taken, t equals twenty seconds. Power, P equals Work done divided by time taken, which is mgh divided by t. This equals four hundred newtons multiplied by eight metres divided by twenty seconds, giving one hundred sixty watts. For girl B: Weight of the girl, mg equals four hundred newtons. Displacement, h equals eight metres. Time taken, t equals fifty seconds. Power, P equals mgh divided by t, which is four hundred newtons multiplied by eight metres divided by fifty seconds, giving sixty-four watts. Power expended by girl A is one hundred sixty watts. Power expended by girl B is sixty-four watts. Example ten point eight. A boy of mass fifty kilograms runs up a staircase of forty-five steps in nine seconds. If the height of each step is fifteen centimetres, find his power. Take g equals ten metres per second squared. Weight of the boy, mg equals fifty kilograms multiplied by ten metres per second squared, giving five hundred newtons. Height of the staircase, h equals forty-five multiplied by fifteen divided by one hundred metres, which is six point seven five metres. Time taken to climb, t equals nine seconds. Power, P equals Work done divided by time taken, which is mgh divided by t. This equals five hundred newtons multiplied by six point seven five metres divided by nine seconds, giving three hundred seventy-five watts. Power is three hundred seventy-five watts. [CHECKPOINT]
Questions for power: One, what is power? Power is the rate of doing work or the rate of transfer of energy. Two, define one watt of power. One watt is the power of an agent which does work at the rate of one joule per second. Three, a lamp consumes one thousand joules of electrical energy in ten seconds. What is its power? Power equals work divided by time, which is one thousand joules divided by ten seconds, giving one hundred watts. Four, define average power. Average power is obtained by dividing the total energy consumed by the total time taken. Activity ten point seventeen. 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 six thirty am and six thirty pm. Do this activity for about a week. How many units are consumed during day time? 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. [CHECKPOINT]
Let us review what you have learnt. Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the object in the direction of the applied force. The unit of work is joule. One joule equals one newton multiplied by one metre. Work done on an object by a force would be zero if the displacement of the object is zero. An object having capability to do work is said to possess energy. Energy has the same unit as that of work. An object in motion possesses what is known as the kinetic energy of the object. An object of mass, m moving with velocity v has a kinetic energy of one half mv squared. The energy possessed by a body due to its change in position or shape is called the potential energy. The gravitational potential energy of an object of mass, m raised through a height, h from the earth surface is given by mgh. 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 constant. Energy exists in nature in several forms such as kinetic energy, potential energy, heat energy, chemical energy etc. The sum of the kinetic and potential energies of an object is called its mechanical energy. Power is defined as the rate of doing work. The SI unit of power is watt. One watt equals one joule per second. [CHECKPOINT]
Now, let us solve the Exercises. Exercise one. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term work. Suma is swimming in a pond. Work is done because she applies force to push water backward and gets displaced forward. A donkey is carrying a load on its back. Work is not done by the donkey on the load because the force is upward while displacement is horizontal, so they are perpendicular. A wind-mill is lifting water from a well. Work is done because force is applied upward and water is displaced upward. A green plant is carrying out photosynthesis. Work is not done in the mechanical sense as there is no force causing displacement of an object. An engine is pulling a train. Work is done because the engine applies a force and the train moves in the direction of the force. Food grains are getting dried in the sun. Work is not done in the mechanical sense. A sailboat is moving due to wind energy. Work is done because wind exerts force on the sails and the boat is displaced. [CHECKPOINT]
Exercise two. 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 net vertical displacement is zero. Gravity acts vertically, and since the object returns to the same horizontal level, the vertical displacement is zero, so work equals force times zero displacement. Exercise three. A battery lights a bulb. Describe the energy changes involved in the process. Chemical energy stored in the battery is converted into electrical energy. The electrical energy flows through the circuit and is converted into light energy and heat energy in the bulb. Exercise four. Certain force acting on a twenty kilogram mass changes its velocity from five metres per second to two metres per second. Calculate the work done by the force. Work done equals change in kinetic energy. Initial kinetic energy is one half times twenty times five squared, which is two hundred fifty joules. Final kinetic energy is one half times twenty times two squared, which is forty joules. Work done equals final minus initial, which is forty minus two hundred fifty, giving minus two hundred ten joules. [CHECKPOINT]
Exercise five. A mass of ten kilograms 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 by gravitational force is zero. Gravitational force acts vertically downward, while the displacement is horizontal. The angle between force and displacement is ninety degrees, and the cosine of ninety degrees is zero, so work done is zero. Exercise six. 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. As potential energy decreases, kinetic energy increases by an equal amount. The total mechanical energy remains constant. Exercise seven. What are the various energy transformations that occur when you are riding a bicycle? Chemical energy from food is converted into muscular energy. Muscular energy is converted into mechanical energy to move the pedals. Mechanical energy is transferred to the wheels as kinetic energy. Some energy is also lost as heat due to friction. [CHECKPOINT]
Exercise eight. 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? No mechanical work is done on the rock because there is no displacement. The energy you spend is converted into heat energy in your muscles and is dissipated to the surroundings. Exercise nine. A certain household has consumed two hundred fifty units of energy during a month. How much energy is this in joules? One unit of electrical energy equals one kilowatt hour, which is three point six times ten to the power of six joules. Two hundred fifty units equals two hundred fifty multiplied by three point six times ten to the power of six joules, which equals nine times ten to the power of eight joules. Exercise ten. An object of mass forty kilograms is raised to a height of five metres 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 equals mgh, which is forty times nine point eight times five, giving one thousand nine hundred sixty joules. At half-way down, height is two point five metres. Potential energy at that point is forty times nine point eight times two point five, giving nine hundred eighty joules. By conservation of energy, kinetic energy equals initial potential energy minus current potential energy, which is one thousand nine hundred sixty minus nine hundred eighty, giving nine hundred eighty joules. [CHECKPOINT]
Exercise eleven. 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 in circular orbit is zero. The gravitational force acts towards the centre of the earth, while the displacement of the satellite is tangential to the orbit. The angle between force and displacement is ninety degrees, so work done is zero. Exercise twelve. 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, an object can have displacement without any net force acting on it. According to Newton's first law, an object moving with constant velocity in a straight line continues to do so without any net external force. Exercise thirteen. A person holds a bundle of hay over his head for thirty minutes and gets tired. Has he done some work or not? Justify your answer. No, he has not done any mechanical work on the bundle of hay. Although he applies an upward force to counteract gravity, there is no displacement of the bundle. Work requires both force and displacement in the direction of force. [CHECKPOINT]
Exercise fourteen. An electric heater is rated one thousand five hundred watts. How much energy does it use in ten hours? Power equals one thousand five hundred watts, which is one point five kilowatts. Time equals ten hours. Energy consumed equals power multiplied by time, which is one point five kilowatts multiplied by ten hours, giving fifteen kilowatt hours or fifteen units. Exercise fifteen. 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 the bob is drawn to one side, it has maximum potential energy and zero kinetic energy. When released, potential energy converts to kinetic energy as it moves down. At the lowest point, kinetic energy is maximum and potential energy is minimum. As it rises on the other side, kinetic energy converts back to potential energy. The bob eventually comes to rest due to air resistance and friction at the pivot. The mechanical energy is gradually converted into heat energy and sound energy, which dissipates into the surroundings. The total energy of the system and surroundings remains constant, so it is not a violation of the law of conservation of energy. [CHECKPOINT]
Exercise sixteen. 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 work done equals the change in kinetic energy. Initial kinetic energy is one half m v squared. Final kinetic energy is zero. Work done equals zero minus one half m v squared, which is minus one half m v squared. Exercise seventeen. Calculate the work required to be done to stop a car of one thousand five hundred kilograms moving at a velocity of sixty kilometres per hour. Convert velocity to metres per second: sixty kilometres per hour equals fifty thirds metres per second. Initial kinetic energy is one half times one thousand five hundred times fifty thirds squared. This equals six hundred twenty-five thousand thirds joules, which is approximately two hundred eight thousand three hundred thirty-three point three three joules. Work required to stop it equals the negative of this initial kinetic energy, so minus two hundred eight thousand three hundred thirty-three point three three joules. The magnitude of work required is two hundred eight thousand three hundred thirty-three point three three joules. Exercise eighteen. 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. In the first diagram, force is perpendicular to displacement, so work done is zero. In the second diagram, force is in the same direction as displacement, so work done is positive. In the third diagram, force is in the opposite direction to displacement, so work done is negative. [CHECKPOINT]
Exercise nineteen. 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 her. If the vector sum of all forces acting on the object is zero, the net force is zero. According to Newton's second law, if net force is zero, acceleration is zero. The object will either remain at rest or move with constant velocity. Exercise twenty. Find the energy in joules consumed in ten hours by four devices of power five hundred watts each. Total power equals four multiplied by five hundred watts, which is two thousand watts or two kilowatts. Time equals ten hours. Energy consumed equals power multiplied by time, which is two kilowatts multiplied by ten hours, giving twenty kilowatt hours. Converting to joules: twenty multiplied by three point six times ten to the power of six joules equals seven point two times ten to the power of seven joules. Exercise twenty-one. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy? When the object hits the ground, its kinetic energy is converted into other forms of energy. It causes sound energy, heat energy due to friction and deformation, and potential energy of deformation if the ground or object is slightly compressed. The kinetic energy is not lost but transformed into these other forms. [CHECKPOINT]
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]