Work and Energy PDF

# Work and Energy ## Physics 5 The weightlifter does no work on the weights as he holds them on his shoulders. He does work only when he raises the weights to this height. 'Work is done on an object when a force causes a displacement of the object'. ## 5.1 Introduction In common usage, the word...

# Work and Energy ## Physics 5 The weightlifter does no work on the weights as he holds them on his shoulders. He does work only when he raises the weights to this height. 'Work is done on an object when a force causes a displacement of the object'. ## 5.1 Introduction In common usage, the word 'work' means any physical or mental exertion. But in physics, work has a distinctly different meaning. Let us consider the following situations: * A student holds a heavy chair at arm's length for several minutes. * A student carries a bucket of water along a horizontal path while walking at constant velocity. * A student applying force against a wall. * A student studying whole day to prepare for examinations. It might surprise you to know that in all the above situations, no work is done according to the definition of work in physics, even though effort is required in all cases. ## 5.2 The concept of work In physics, the word 'work' has a definite and precise meaning. Imagine that your car has run out of fuel and you have to push it down the road to the gas station. Let your friend push the car with a constant horizontal force. If the car does not move, no work is done. Suppose, he increases the magnitude of this force by pushing the car harder. If the car starts moving, he does a work on the car. A person exerts a force on the car and displaces it to the left. The work done is done by him on the car. ## CBSE: Class IX Work is not done on an object unless the object is moved with the action of a force. The application of a force alone does not constitute work. For example, when a student holds the chair in his hand, he exerts a force to support the chair. But, work is not done on the chair as the chair does not move. Some more examples to understand the concept of work are given below: * Push a box lying on a surface. The box moves through a distance. You exerted a force on the box and the box got displaced. In this situation work is done. * A boy pulls a trolley and the trolley moves through a distance. The boy has exerted a force on the trolley and it is displaced. Therefore, work is done. * A person lifts a cat through a certain height. To do this he applies a force. The cat rises up. There is a force applied on the cat and the cat has moved. Hence, work is done. Two important conditions that must be satisfied for work to be done are: * a force should act on an object * the object must be displaced. * direction of force must not be perpendicular to the displacement. If any one of the above conditions does not exist, work is not done. This is the concept of work that we use in science. ## BUILDING CONCEPTS 5.1 A horse is pulling a cart. The cart moves. Do you think that work is done in this situation? **Explanation** Yes, in this situation a work is done. When a horse pulls a cart, it is applying a force that moves the cart. Since a force is applied on the cart and the cart is displaced, a work is done by the horse on the cart. ## 5.3 Mathematical definition of work * **A constant force is applied in the direction of the displacement of an object:** Let a constant force, Facts on an object. Let the object be displaced through a distance, s in the direction of the force. Let W be the work done. Here, we define work to be equal to 'the product of the force and displacement'. Work done = force x displacement W=Fxs * **A constant force is applied at a certain angle with the direction of the displacement of an object:** When the force on an object and the object's displacement are in different directions, the work done on the object is given by, W=Fx sx cosθ where, the angle between the force and the direction of the displacement is 0 (see fig.5). Here, we define work to be equal to 'the force multiplied by the displacement multiplied by the cosine of the angle between them'. Work is a scalar quantity; it has only magnitude and no direction. 1 Joule = 1 newton x 1 meter SI unit of work: Joule or 1 J=1Nm **Definition of 1 joule:** 1 J is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force. **Some important points related to work** * If 0 = 0°, then cos 0° = 1 and W = F x s. * If 0 = 90°, then, W = 0 because cos 90° = 0. So, no work is done on a bucket being carried by a girl walking horizontally (see fig. 6). The upward force exerted by the girl to support the bucket is perpendicular to the displacement of the bucket, which results in no work done on the bucket. * If 0 = 180°, then cos 180° = -1 and W = - Fx s. ## 5.4 Concept of negative and positive work The work done by a force can be either positive or negative. * **Whenever angle (0) between the force and the displacement is acute, i.e., 0° <0 < 90°, the work done is positive. Also, when angle (0) between the force and displacement is zero, i.e., force and displacement are in same direction, the work done is positive.** * **Whenever angle (0) between the force and the displacement is obtuse, i.e., 90° <0 < 180°, the work done is negative. Also, when angle (0) between the force and displacement is 180°, i.e., force and displacement are in opposite direction, the work done is negative.** ## NUMERICAL ABILITY 5.1 A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force (see fig.8). If the force acts on the object all through the displacement, then what is the work done on the object? **Solution** Given, force, F = 5 N; displacement, s = 2 m; W = ? Then, the work done, W = Fxs=5x2 = 10 J. ## NUMERICAL ABILITY 5.2 How much work is done on a vacuum cleaner pulled 3.0 m by a force of 50.0 N at an angle of 30.0° with the horizontal? **Solution** Given, F = 50.0 N; 0 = 30.0°; s = 3.0 m; W = ? Work done, W = Fs cos0 = 50 x3 x cos 30° = 50 x3 x √3/2 or W = 75√3=75 ×1.73 = 129.75 J ## ACTIVE PHYSICS 5.1 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. (i) Which one of these forces is doing positive work? (ii) Which one is doing negative work? Give reasons. **Explanation** The force applied by you (Fapplied) is in upward direction while the force of gravity or weight of the body (W= mg) is in downward direction (see fig. 9). (i) Now, displacement (s) done by you is in upward direction. Since the applied force (Fapplied) and the displacement (s) are both in same direction, the work done by applied force is positive. (ii) Now, force of gravity (W) and the displacement (s) are in opposite direction, thus, the work done by the gravity is negative. ## BUILDING CONCEPTS 5.2 An artificial satellite is moving around the Earth in a circular path under the influence of centripetal force provided by the gravitational force between them. What is the work done by this centripetal force? **Explanation** Centripetal force (F) is always perpendicular to the displacement (s) of the particle moving along a circular path. That is, the angle (0) between them is 90°. Now, work done, W = F s cos 0 = F s cos 90° = 0 [cos 90° = 0] Thus, work done by this centripetal force is zero. Work done by the centripetal force is always zero because it is always perpendicular to the displacement. For example, if an electron moves around a nucleus in a circular path due to centripetal force provided by the electric force between them, the work done by this force is zero. ## 5.5 Work done by applied force against gravity If an object is lifted up to a certain height (see fig. 11), definitely, a work is done by the applied force. The applied force must be equal to the weight ( = mg) of the object. This work done is given by, W = F xs = (mg) xh Where, m = mass of object; g = acceleration due to gravity; h = height or W = mgh ## NUMERICAL ABILITY 5.3 A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage (Take g = 10 m/s²). **Solution** Given, mass of luggage, m = 15 kg; height, h = 1.5 m ; acceleration due to gravity, g = 10 m/s²; W = ? Work done, W = mgh = 15 x 10 x 1.5 m = 225 J ## CHECK YOUR CONCEPTS 5.1 * A woman walks along with a constant velocity holding a suitcase. How much work is done by the force holding the case? * The gravitational force exerted by the Sun on the Earth holds the Earth in an orbit around the Sun. Let us assume that the orbit is perfectly circular. The work done by this gravitational force during a short time interval in which the Earth moves through a displacement in its orbital path is (a) zero (b) positive (c) negative (d) impossible to determine. * Figure 12 shows four situations in which a force is applied to an object. In all four cases, the force has the same magnitude, and the displacement of the object is to the right and of the same magnitude. Rank the situations in order of the work done by the force on the object, from most positive to most negative. ## 5.6 Energy Without light that come to us from the Sun, life on Earth would not exist. With the light energy, plants can grow and the oceans and atmosphere can maintain temperature ranges that support life. Although energy is difficult to define comprehensively, a simple definition is that energy is the capacity to do work. Thus, when you think of energy, think of what work is involved. Let us take some examples: (i) Energy must be supplied to a car's engine in order for the engine to do work in moving the car. In this case, the energy may come from burning petrol. (ii) When a fast moving cricket ball hits a stationary wicket, the wicket is thrown away. Thus, work is done by the energy in the moving ball on the wicket. (iii) An object placed at a certain height has the capability to do work. If it is allowed to fall, it will move downward i.e., a work will be done in this case. (iv) When a raised hammer falls on a nail placed on a piece of wood, it drives the nail into the wood. Thus, energy of hammer does a work on nail. (v) When a balloon is filled with air and we press it we notice a change in its shape. That is, we have done a work on balloon to change its shape. As long as we press it gently, it can come back to its original shape when the force is withdrawn. This means, balloon acquires energy due to which it regains its original shape. (vi) If we press the balloon hard, it can even explode producing a blasting sound. Again, it acquires enough energy that does work to blast the balloon. In all the above examples, the objects acquire the capability of doing work which is called energy. ## BUILDING CONCEPTS 5.3 How does an object with energy do work? **Explanation** An object that possesses energy can exert a force on another object. When this happens, energy is transferred from first object to the second object. 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. **SI unit of energy:** Since energy is the capacity to do work, its unit is same as that of work, that is, joule (J). 1 J is the energy required to do 1 joule of work. Sometimes a larger unit of energy called kilo joule (kJ) is used, 1 kJ = 1000 J. ## 5.7 Forms of energy The world we live in provides energy in many different forms. The various forms include potential energy, kinetic energy, heat energy, chemical energy, electrical energy and light energy. ## Mechanical energy The capacity to do mechanical work is called mechanical energy. Mechanical energy can be of two types: * Kinetic energy * Potential energy The sum of the gravitational potential energy and the kinetic energy is called mechanical energy. ## Kinetic energy This is the energy a body has due to its movement. To give a body KE, work must be done on the body. The amount of work done will be equal to the increase in KE. Kinetic energy is the energy associated with an object in motion. **Derivation of kinetic energy** Let a constant force F acts on a ball of mass m with an initial velocity u. The displacement of the ball be 's', time taken to displace it be 't', its final velocity be 'v' and acceleration produced in it be 'a' (see fig. 13). Now, work done, W = F xs or W = (ma) xs (F = m xa) Now, from third equation of motion, we have, v² = u² + 2as or a = (v² - u²)/2s (2) or W = mx((v² - u²)/2s) ×s or W = (mv²-mu²)/2 (3) It is clear from equation 3, the work done is equal to the change in the kinetic energy of an object. Equation 3 is called Work-Energy Theorem. If u = 0, the work done will be W = (mv²-m(0)²)/2 = 1/2mv² Thus, the kinetic energy possessed by an object of mass, m and moving with a uniform velocity, v is Ek=1/2mv² If the speed of an object is doubled, its kinetic energy becomes four times the initial value. This is because kinetic energy is directly proportional to the square of the speed of the object. If the mass of an object in motion is doubled, then its kinetic energy is also doubled, as kinetic energy is directly proportional to the mass of the object. ## NUMERICAL ABILITY 5.4 An object of mass 15 kg is moving with a uniform velocity of 4 m s-¹. What is the kinetic energy possessed by the object? **Solution** Given, mass of the object, m = 15 kg; velocity of the object, v = 4 m s-¹ ; kinetic energy, E₁ = ? Ex = 1/2mv² = 1/2 ×15×(4)² = 120 J ## NUMERICAL ABILITY 5.5 What is the work to be done to increase the velocity of a car from 30 km h-¹ to 60 km h-¹ if the mass of the car is 1500 kg? **Solution** Given, mass of the car, m = 1500 kg; initial velocity of car, u = 30 km h-¹ = 30x5/18 m s¯¹; final velocity of the car, v = 60 km h-1 = 60x5/18 ms-1; Now, work done = change in kinetic energy = EKF - EKI where, Ekt = final kinetic energy and E₁₁ = initial kinetic energy W=1/2mv²-1/2mu² = m(v²-u²) = 1/2×1500( (50/3)²-(25/3)²) or W = 1/2 ×1500( 2500/9 - 625/9 = 1/2×1500 1875/9 = 156250.1 J ## NUMERICAL ABILITY 5.6 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 = 9.8 m s-2. **Solution** Given, mass of the object, m = 10 kg; displacement (height), h = 6 m; acceleration due to gravity, g = 9.8 m s-2. Potential energy = mgh = 10 kg x 9.8 m s-2 x 6 m = 588 J. ## Gravitational potential energy depends on height from an arbitrary zero level It is important to note that the height, h, is measured from an arbitrary zero level. In the example of the egg, if the floor is the zero level, then h is the height of the table, and mgh is the gravitational potential energy relative to the floor. Alternatively, if the table is the zero level, then h is zero. Thus, the potential energy associated with the egg relative to the table is zero. Let us take another example: Suppose you drop a volleyball from a second-floor roof and it lands on the first-floor roof of an adjacent building (see fig. 19). If the height is measured from the ground, the gravitational potential energy is not zero because the ball is still above the ground. But if the height is measured from the first floor roof, the potential energy is zero when the ball lands on the roof. If B is the zero level, then all the gravitational potential energy is converted to kinetic energy as the ball falls from A to B. If C is the zero level, then only part of the total gravitational potential energy is converted to kinetic energy during the fall from A to B. ## NUMERICAL ABILITY 5.7 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 = 10 m s-2. **Solution** Given, mass of the object, m = 12 kg, potential energy, E, = 480 J. Now, E, = mgh or 480 = 12 x 10 xh or h = 480/12×10 = 4 m The object is at the height of 4 m. ## Elastic potential energy Imagine you are playing with a spring on a tabletop. You push a block into the spring, compressing the spring, and then release the block. The block slides across the tabletop. The kinetic energy of the block came from the stored energy in the compressed spring (see fig. 20 on next page). This potential energy is called elastic potential energy. Elastic potential energy is stored in any compressed or stretched object, such as a spring or the stretched strings of a tennis racket or guitar. The length of a spring when no external forces are acting on it is called the relaxed length of the spring. When an external force compresses or stretches the spring, elastic potential energy is stored in the spring. The amount of energy depends on the distance the spring is compressed or stretched from its relaxed length. ## CHECK YOUR CONCEPTS 5.2 * Is it possible for a slow moving elephant to have more kinetic energy than a fast moving fox? * Is it possible for kinetic energy of an object to be negative ? * A sailboat is moving at a constant velocity. What is the work done by net external force acting on the boat ? ## ACTIVE PHYSICS 5.4 * Take a toy car. Wind it using its key (see fig. 21). Place the car on the ground. You will find that the car will move. * When the key of the car is wound, its spring is coiled and thus, an elastic potential energy is stored in it. Now, give three or four windings to the key and then allow the car to move along the ground. Note down the distance it has covered before it stops. Next, give seven or eight windings to the key and again allow the car to move along the ground. You will find that this time the distance covered by the car is more than before. This shows that more the number of windings of the key, more will be the elastic energy stored in the spring. ## ACTIVE PHYSICS 5.5 * Take a long spring as shown in fig. 22. Ask a friend to hold one of its ends. You hold the other end and move away from your friend. * Now you release the spring. You will observe that the spring tends to regain its original configuration. This is because an elastic potential energy is stored in it on account of work done in stretching it. * If the spring is compressed, still the spring will acquire an elastic potential energy as a result of work done on it. ## ACTIVE PHYSICS 5.6 * Take a bow as shown in fig. 23. 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. You will notice that the arrow flies off the bow. Also, the bow regains its original shape (or configuration). * Here, the potential energy stored in the bow due to the change of shape (or configuration) is used in the form of kinetic energy in throwing off the arrow. * If the string is stretched more, the bow bends more, i.e., more will be the elastic potential energy stored in it. When the string is released, the bow now imparts a large kinetic energy to the arrow, thus, the arrow travels a greater distance. ## 5.8 The law of conservation of energy Energy appears in many forms, such as heat, motion, height, pressure, electricity, and chemical bonds between atoms. ## Energy transformations Energy can be converted from one form to another form in different systems, machines or devices. Systems change as energy flows from one part of the system to another. Parts of the system may speed up, slow down, get warmer or colder, etc. Each change transfers energy or transforms energy from one form to another. For example, friction transforms energy of motion to energy of heat. A bow and arrow transform potential energy in a stretched bow into energy of motion (i.e., kinetic energy) of an arrow. ## Law of conservation of energy Energy can never be created or destroyed, just converted from one form into another. This is called the law of conservation of energy. The law of conservation of energy is one of the most important laws in physics. It applies to all forms of energy. ## Energy has to come from somewhere The law of conservation of energy tells us that energy cannot be created from nothing. If energy increases somewhere, it must decrease somewhere else. The key to understanding how systems change is to trace the flow of energy. Once we know how energy flows and transforms, we have a good understanding of how a system works. For example, when we use energy to drive a car, that energy comes from chemical energy stored in petrol. As we use the energy, the amount left in the form of petrol decreases. ## Conservation of mechanical energy The mechanical energy i.e., the sum of potential and kinetic energies is constant in the absence of any frictional forces. This means that if you calculate the mechanical energy (E) at any two randomly chosen times, the answers must be equal. Let us take an example of free fall (here, the effect of air resistance on the motion of the object is ignored) : Let us consider an object of mass 'm' at a certain height 'h' (see fig. 24). Let it is dropped from this height from point A i.e., v₁ = 0. Let after a certain time 't', it reaches point B after covering a distance 'x'. Now, from third equation of motion, we have, Mechanical energy at B, E₃ = K.E. at B + P.E. at B = 1/2mvB² + mghB or Eg = 1/2m(2gx) + mg(h - x) = mgx + mgh-mgx = mgh Finally, the ball reaches the ground (at point C) after covering a distance h. Again, from third equation of motion, we have, v2 = VA 2 + 2gh = (0)2 + 2gh Mechanical energy at C, Ec = K.E. at C + P.E. at C = 1/2mvc² + mghc or Ec = 1/2m(2gh) + mg(0) = mgh From eqs.(1), (3) and (5), we get, EA = EB = Ec This means total mechanical energy is conserved during the free fall of an object. That is, 1/2mv² + mgh = constant ## CHECK YOUR ANSWERS 5.2 * Yes, the kinetic energy of a slow moving elephant may be more than kinetic energy of a fast moving fox. This is because the mass of elephant is very large as compared to mass of fox. * Kinetic energy (=1/2mv²) is always positive as mass (m) and square of speed (v²) both are positive. * We know that net work done by all the forces acting on an object is equal to the change in the object's kinetic energy. Since, the sailboat is moving with constant velocity, the change in kinetic energy of sailboat is zero. Thus, the work done by the net external force acting on the boat is zero. ## ACTIVE PHYSICS 5.7 Discuss following questions in your group: * 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? ## NUMERICAL ABILITY 5.8 A 4.0 x104 kg roller coaster starts from rest at point A. Neglecting friction, calculate its potential energy relative to the ground, its kinetic energy, and its speed at point B (see fig.27). Take g = 10 m/s². **Solution** Initial energy (at A), or EA = 1/2mv+mgha = 1/2m(0)² + mg(54) EA = 4.0 x 104 x 10 x 54 = 2.16 × 107 J Potential energy at B, EPB = mxgxh₁ = 4.0 x104 ×10×15 = 6 x 106 J Kinetic energy at B, Екв = ЕА - Ерв = 2.16 x 107-6 x 106 = 15.6 x 106 J NOW, EKB =MVB²/2 or VB²= 2E.KB/m or VB=√(2×15.6×10⁶/4.0×10⁴) = √780 = 27.9 m/s ## 5.9 Power The engine in an old school bus could, over a long period of time, do as much work as jet engines do when a jet takes off. However, the school bus engine could not begin to do work fast enough to make a jet lift off. In this and many other applications, the rate at which work is done is more significant than the amount of work done. **Power is the rate at which work is done. Power can also be defined as the rate at which energy is transferred.** P = Work done/time taken **SI unit of power: Watt (W).** or 1 W=1Js-1 **Definition of 1 watt:** If 1 joule work is done per second by a device or a machine, then the power of that device or machine is 1 watt. ## NUMERICAL ABILITY 5.9 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** (i) Power expended by girl A: Weight of the girl, mg = 400 N; displacement (height), h = 8 m; time taken, t = 20 s P = Work done/time taken = mgh/t = 400×8/20 = 160 W (ii) Power expended by girl B: Weight of the girl, mg = 400 N displacement (height), h = 8 m; time taken, t = 50 s P = Work done/time taken = mgh/t = 400×8/50 = 64 W ## NUMERICAL ABILITY 5.10 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 = 10 m s-2. **Solution** Weight of the boy, W = mg = 50 x 10 = 500 N; height of the staircase, h = 45 x 15 cm = (45 x 15)/100 m = 6.75 m ; time taken to climb, t = 9 s P = Work done/time taken = mgh/t = 500×6.75/9 = 375 W ## Power in terms of force (F) and velocity (v) We know that power, P = W/t = Fs/t = F(s/t) or P=Fxv ## 5.10 Commercial unit of energy The unit joule is an extremely small unit. It is inconvenient to express large quantities of energy in terms of joule. We use a bigger unit of energy called kilowatt hour (kWh). It is called commercial unit of energy. **Definition of 1 kWh:** If a machine or a device of power 1 kW or 1000 W is used continuously for one hour, it will consume 1 kW h of energy. Thus, 1 kWh is the energy used in one hour at the rate of 1000 W (or 1 kW). or 1 kWh = 1 kW x1h = 1000 W ×3600 s = 3600000 J 1 kWh = 3.6 ×106 J. ## NUMERICAL ABILITY 5.11 An electric bulb of 60 W is used for 6 h per day. Calculate the 'units' of energy consumed in one day by the bulb. **Solution** Power of electric bulb = 60 W = 0.06 kW; time used, t = 6h Energy = power xtime taken = 0.06 kW x6 h = 0.36 kWh = 0.36 'units'. The energy consumed by the bulb is 0.36 'units'. ## BUILDING CONCEPTS 5.4 A light body and a heavy body have same kinetic energy. Which one of the two has greater momentum? **Explanation** Firstly, we will find the relationship between kinetic energy and linear momentum. Kinetic energy, Ek = 1/2mv² = (1/2)mv² x (m/m) = (mv)²/2m or Ek = p²/2m (p= mv = linear momentum) or p² = 2mEx or p = √2mEK This means, if kinetic energy (ER) is constant for both the bodies, then, p√m. Thus, heavier body will have greater momentum than the lighter body. ## CHECK YOUR ANSWERS 5.1 * No work is done by the force holding the suitcase while walking. This is because force (F) and the displacement (s) are perpendicular to each other (see fig. 14). * Option (a) is correct i.e., work done is zero as force (F) and displacement (s) are perpendicular to each other. * Ranks starting from most positive to most negative are as follows: (c), (a), (d), (b). ## ACTIVE PHYSICS 5.2 * 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. You will find that more the height from which the ball is dropped, more will be depth of the depression formed by it in the wet bed of sand. This is because more the height from which the ball is dropped, more will be the speed with which it strikes the wet bed. This means more will be its kinetic energy and it will do more work i.e., will penetrate to a greater depth. ## ACTIVE PHYSICS 5.3 * Set up the apparatus as shown in fig. 15. 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. * The pan along with mass falls due to gravity. Since, trolley is attached to the pan, it also starts moving i.e., it gains kinetic energy. Now, when the moving trolley hits the wooden block, a portion of kinetic energy of the trolley is transferred to wooden block. Due to this the block moves and gets displaced to a certain distance. * Repeat this activity by increasing the mass on the pan. More the mass of the pan, more will be the kinetic energy gained by trolley and hence, more will be the kinetic energy gained by the wooden block. Thus, more will be its displacement. We can conclude that 'a moving object can do work'. * An object moving faster can do more work than an identical object moving relatively slow. ## Potential energy The energy possessed by an object due to its position or configuration is called 'potential energy'. Consider the balanced smooth rock shown in fig. 16. As long as the rock remains balanced, it has no kinetic energy. If it becomes unbalanced, it will fall vertically to the ground and will gain kinetic energy as it falls. The origin of this kinetic energy is potential energy present in the rock. Thus, potential energy is stored energy. Potential energy is associated with an object that has the potential to move because of its position or configuration. ## Gravitational potential energy The energy associated with an object due to the object's position relative to a gravitational source is called gravitational potential energy. Gravitational potential energy is energy due to an object's position in a gravitational field. Imagine an egg falling off a table. As it falls, it gains kinetic energy. But, where does the egg's kinetic energy come from? It comes from the gravitational potential energy that is associated with the egg's initial position on the table relative to the floor. ## Derivation of 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 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'. Let us consider an object of mass, m which is raised through a height, h from the ground (see fig. 17). A force equal to the weight (= mg) of the object is required to do this. The object gains energy equal to the work done on it. Work done on the object, W = force xdisplacement = mg ×h = mgh This work done on the object is the energy gained by