6 WORK and ENERGY. 6.0 Introduction. 6.1 Work and kinetic energy. Objectives

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1 6 WORK and ENERGY Chapter 6 Work and Energy Objectives After studying this chapter you should be able to calculate work done by a force; be able to calculate kinetic energy; be able to calculate power; be able to use these quantities in solving problems; be able to model problems involving elastic strings and springs; know when mechanical energy is conserved. 6.0 Introduction When you walk, run or cycle up a hill or go upstairs you know that it takes more effort than on the level. You may take part in gymnastics or skiing, you may play squash or go down the water chute at the local Leisure Centre, or you may go on rides like the log flume or the Corkscrew at the Amusement Park. All these activities involve the ideas of work, energy and power developed in this chapter. 6.1 Work and kinetic energy Example To move a car of mass 1000 kg along a level road you need to apply a force of about 200 N in the direction of motion to overcome the resistance and get going. If this resistance is 100 N and you push it 20 m, what is the speed of the car? at rest 200 N 100 N 20 m v Solution The net force on the car is ( )N =100 N, so if a is the constant acceleration, Newton's Second Law gives 100 = 1000a (1) giving a = = 1 10 ms 2 133

2 Since the car is initially at rest, if v is its speed after being pushed 20 m or v 2 = 0 + 2a 20 = 40a = 4 v = 2ms 1. (The result v 2 = u 2 + 2as for constant acceleration was introduced in Chapter 2.) Consider next the general case of a car of mass m kg. If P is the force given to the car by the person pushing it in the direction of motion, and R the resistance then the equation of motion replacing (1) is so that P R = ma a = ( P R) m. (2) When the car has moved a distance x metres from rest its speed v is given by or rearranging v 2 = 2ax = 2 ( P R) x, using equation (2), m ( P R)x = 1 2 mv2. (3) P RF The right hand side of equation (3) depends upon the mass and the final speed of the car. It represents the energy of the car due to its motion and is called the kinetic energy. Kinetic energy = 1 2 (mass) (speed)2 (4) = 1 2 mv2 The car gains energy as it speeds up. Where has this energy come from? It comes from you pushing the car. To move the car you do work. Some of this work overcomes the resistance and the rest makes the car accelerate. Equation (3) expresses these physical facts mathematically. Px is called the work done by the force P. This is the work you do in pushing the car. 134

3 Work done by a constant force = (Force) (distance moved in the direction of the force) The term Rx in equation (3) is the work done by the resistance. The negative sign occurs since the direction of the force is opposite to the direction of motion. If the work done by a force is negative, work is said to be done against the force. The left hand side of equation (3) is the work done by the resultant force in the direction of motion and is equal to (work done by the force P) (work done against the force R) It is this positive work done which produces the kinetic energy. This concept of work is a special case of what is meant by the term 'work' in everyday language. To do work in mechanics an object has to move through some distance. The athlete moving the weights vertically upwards does work against the force of gravity. (Gravity is acting downwards, the weights are being moved upwards). To measure both kinetic energy and the work done by a force we need some units and from equation (3) the units of both these quantities will be the same. The unit of work is the joule (J), so that 1 J is the work done in moving a force of 1 N through 1 m. Thus 1J =1Nm. For the weights of mass 100 kg being lifted 2 m the work done against gravity is 100 g 2 = J taking g = 10 ms 2. Since the left hand side of equation (3) represents work done and the right hand side kinetic energy, equation (3) is called the workenergy equation. The jumping flea The flea Spilopsyllus weighs about 0.4 mg and leaves the ground, when it jumps, at about 1 ms 1. What is its kinetic energy? 135

4 In this example the unit of mass needs to be changed to kg. Now and 0.4mg= kg, kinetic energy = 1 ( ) 1 () 2 = = J. How is this energy used? Can you say anything about how high the flea jumps? Why does the earthworm move more slowly than the centipede? Kinetic energy changes are involved here too. When an earthworm, which has no legs, crawls each segment moves in steps coming to a halt between each step. The whole body is continually being given kinetic energy by the muscles doing work and this energy is quickly dissipated at every stage. For this reason the worm only crawls at about 0.5 c ms 1. On the other hand, a fast species of centipede only 2.2 cm long can run on its legs at 0.45 ms 1. Although at each step the kinetic energy supplied to its legs fluctuates their mass is small compared to the total mass of the centipede and so the main part of the body, which keeps moving, moves at an almost constant speed and so has almost constant kinetic energy, enabling the centipede to move quickly. Activity 1 Gravel beds These are often to be found at the side of some holiday routes and on bendy stretches of road in, for example, the Peak District National Park Their purpose is to bring a vehicle to a halt safely in the event of brake failure or cornering too fast. The car enters the bed at some speed and so possesses some kinetic energy. It is brought to rest without braking by the friction of the bed. As it slows down it loses energy. What happens to this energy? Suppose a car of mass 1 tonne enters the bed at 80 kph and stops after 30 m. Find how much kinetic energy is lost and calculate the average resistance force produced by the gravel. 136

5 As a simplified model suppose that a car of mass m kg enters the bed with speed v ms 1 and comes to rest after x metres. Suppose that the bed is horizontal and that the resistance force, R, is constant. Find the deceleration in terms of R and m and hence derive the work-energy equation Rx = 1 2 mv 2. RF x Example Green to amber A car of mass 1300 kg is travelling along a straight level road at 10 ms 1 when the traffic lights change from green to amber. The driver applies the brakes 23 metres from the lights and just manages to stop on the line. Calculate (a) the kinetic energy of the car before braking (b) the work done in bringing the car to rest (c) the force due to the brakes, friction, etc. against which work is done. Solution (a) The kinetic energy of the car before braking = ( 10 )2 10 ms m = J. (b) Using equation (3), work done in bringing car to rest = kinetic energy lost = J. (c) Since work = (force) (distance) force = work distance = = 2826 N. 23 Part (b) of this example illustrates how the work-energy equation can be used to calculate the work done without knowing the forces involved explicitly. Normal contact force Returning to the model of the gravel bed and also the earlier example of pushing the car, two other forces act on the car (i) (ii) its weight mg, the normal contact force. mg The distance the car moves in the direction of either of these forces is zero and so these forces do no work. 137

6 In general, if the direction of motion is perpendicular to the force, no work is done by the force. The idea of work can be explored a little further. Car on icy slope The car of mass 1300 kg now slides 100 m down an icy slope, the slope being inclined at 30 to the horizontal. If the frictional forces on the slope and air resistance are ignored the forces acting on the car are Normal contact force 30 o mg (i) (ii) its weight 1300g N the normal contact force. The distance the car moves in the direction of the normal contact force is zero and so this force does no work. The only force which does work is the weight. The distance moved in the direction of the weight is 100 sin 30 =50 m 30 o and so work done by gravity = 1300 g 50 = = J. In this calculation the distance is found in the direction of the force. Alternatively the weight can be replaced by its components 1300gsin 30 down the slope 1300g sin30 30 o 1300g cos g cos30 perpendicular to the slope. The component perpendicular to the slope does no work since it is perpendicular to the motion. The only force which does work is the component down the plane and so the work done = ( 1300 g sin 30 ) (distance down the plane) In summary, = 1300 g sin = J as before. Work done by a force = (component of force in the direction of motion) (distance moved). 138

7 Activity 2 Pulling a sledge A girl is pulling a sledge through the snow. What forces act on the sledge? Which of these forces does the work? What happens if the sledge is on a slope? Exercise 6A (Take g = 10 ms 2 ) 1. Find the kinetic energy of the following: (a) a runner of mass 50 kg running at 6 ms 1 (b) a hovercraft of mass 10 5 kg travelling at 25 ms 1 (c) an electron of mass g, speed ms 1 (d) a car of mass 1300 kg travelling at 60 kph. 2. A boy of mass 20 kg rides a bicycle of mass 15kg along a road at a speed of 5 ms 1. He applies his brakes and halves his speed; calculate the reduction in the kinetic energy. 3. A bullet of mass 0.02 kg travelling with speed 100 ms 1 comes to rest when it has gone 0.4 m into sand. Find the resisting force exerted by the sand. 4. A boy of mass 40 kg slides down a chute inclined at 30 to the horizontal. If the chute is smooth and the boy starts from rest, with what velocity does he pass a point 5 m from the starting point? 5. A girl of mass 40 kg slides down the water chute 4 m long at the local leisure centre. If the resultant force down the chute is 200 N with what speed does she leave the chute if she starts from rest? 6. If two men push a car of mass 1000 kg, one at each rear corner, and each exerts a force of 120 N at an angle of 20 to the direction of motion, calculate the work done by the two men in pushing the car 20 m. Assuming that the resistance is 100 N, calculate the speed of the car if it starts from rest. 7. A car of mass 1200 kg travelling at 15 ms 1 comes to rest by braking in a distance of 50 m. If the additional resistance to motion is 100 N calculate the braking force of the car assuming that it is constant. 8. A cable car of mass 1200 kg moves 2 km up a slope inclined at 30 to the horizontal at a constant speed. If the resistance to motion is 400 N find the work done by the tension in the cable. T 30 o 9. A man pushes a lawnmower of mass 30 kg with a force of 30 N at an angle of 35 to the horizontal for 5m. The mower starts from rest and there is a resistance to motion of 10 N. Find the work done by the man and the final speed of the mower. 10. A car of mass 1000 kg travelling at 12 ms 1 up a slope inclined at 20 to the horizontal stops in a distance of 25 m. Determine the frictional force which must be supplied. 11. A 70kg crate is released from rest on a slope inclined at 30 to the horizontal. Determine its speed after it slides 10 m down the slope if the coefficient of friction between the crate and the slope is

8 6.2 Work done against gravity: gravitational potential energy When weights of mass 100 kg are lifted 2 m work done against gravity = 100 g 2J = ( mass) g ( vertical height). Generalising this to the case of weights of mass m kg raised vertically through a distance h m the work done against gravity is mass g vertical height. mgh is the work done against gravity. In the example of the car sliding down the icy slope work done by gravity = 1300 g 50 J If the car had been moving up the slope work done by gravity = 1300 g 50 J = ( mass) g (vertical distance through which the car moved) = ( mass) g (vertical distance through which the car moved) 30 o 1300g 50 or alternatively, work done against gravity = 1300 g 50 J = ( mass) g (vertical distance through which the car moved) For a car of mass m the work done against gravity is still mgh. Since this result does not involve the angle of the slope, θ, it is the same for any slope. If you go upstairs is the work done against gravity still your weight times the vertical height? In fact this result is true for any path joining two points A and B. The work against gravity is always mg (vertical height from A to B), and is independent of the path joining A to B. A 30 o A mg B B h h 140

9 In raising a body up through a height h it has gained energy due to its change in position. Energy due to position is called potential energy. This particular form of potential energy is called gravitational potential energy. Gain in gravitational potential energy in moving an object from A to B = work done against gravity in moving from A to B = (mass) g (vertical height from A to B). 6.3 Work-energy equation. Conservation of energy For the earlier example of pushing a car the work-energy equation is Fx = 1 2 mv2, (5) where F = P R is the resultant force in the direction of motion. How is this equation modified if the car moves with constant acceleration, a, from a speed u to speed v over a distance of x metres? If F is the resultant force in the positive x-direction the equation of motion is still at rest P u FR x v v giving F = ma x FR a = F m. (6) From Chapter 2, you know that v 2 = u 2 + 2ax = u F x, using (6) m or Rearranging mv 2 = mu 2 + 2Fx Fx = 1 2 mv2 1 2 mu2 (7) 141

10 so that work done by the force F = final kinetic energy initial kinetic energy work done by the force F = change in kinetic energy produced by the force. Equation (7) is the work-energy equation for this situation. It reduces to equation (5) in the special case when u = 0. For a body falling under gravity from A to B, F = mg and using equation (7), mgh = 1 2 mv2 1 2 mu2 (8) or work done by gravity = change in kinetic energy produced by gravity. h A B u v Activity 3 Introducing conservation of energy In equation (8) the ball falls a height h. Suppose now you choose a horizontal level so that A is at a height h 1 and B is at a height h 2 above the chosen level. Write down the work done by the force of gravity as the body falls from A to B in terms of h 1 and h 2. Show that the sum of the kinetic energy and potential energy is the same at A and at B. Repeat the calculation with a different horizontal level. What do you find? Defining mechanical energy = kinetic energy + potential energy Activity 3 shows that mechanical energy has the same value at A and B, in other words, mechanical energy is conserved. Activity 3 also shows that it does not matter which horizontal or zero you work from in calculating the potential energy since it is only changes in potential energy that are involved. Once you have chosen the zero level, which can often be done to simplify calculations, you must use the same level throughout. 142

11 When you release a ball from rest, as the height of the ball decreases its potential energy decreases but the speed of the ball increases and so its kinetic energy increases. If air resistance is neglected the only force acting is gravity and from equation (8) putting u = 0 mgh = 1 2 mv2, So the decrease in potential energy = increase in kinetic energy. Since m cancels h at rest v v 2 = 2gh or v = 2gh. So, the speed v can be found from considering energy; the speed v is independent of the mass of the ball; it only depends upon the height from which the ball is released. Does this result hold for a particle sliding down a smooth inclined plane from a vertical height h? As the ball falls its potential energy changes to kinetic energy in such a way that at any time the total energy remains the same. If the ball is lifted a vertical height h and then released from rest, the work done against gravity in lifting the ball can be recovered as kinetic energy and for this reason the force of gravity is called a conservative force. A property of such forces is that the work done is independent of the path. On the other hand, forces such as friction and air resistance are not conservative because the work done against them is dissipated and so cannot be recovered. It appears mainly in the form of heat. h v Activity 4 Dropping a ball from rest You will need a bouncy ball, 2 metre rulers, and scales to weigh the ball for this activity. Drop a ball from 2 m above a hard floor surface, releasing the ball from rest. What happens to the ball? Explain the motion in terms of energy. Calculate the speed of the ball when it hits the floor. Why does the ball not bounce back to the same height? Measure how high the ball bounces. Calculate the loss of mechanical energy. If there is a loss of mechanical energy, where does this energy go? 143

12 Example Two masses 4 kg and 6 kg are attached to the ends of an inextensible string over a pulley. The weights of the string and pulley are much smaller than the two masses attached and so are neglected. The pulley is also assumed to be smooth so that the tension T is the same on both sides of the pulley. The masses are held at the same level and the system is released from rest. What is the speed of the 6 kg mass when it has fallen 2 metres? T T T T Solution T Equation (7) can be applied to the two masses separately. For the 4 kg mass the resultant force in the direction of motion is T 4g and so when this mass has risen 2 metres, T T 4g T or ( T 4g) distance = change in kinetic energy 4g 6g 6g ( T 4g) 2 = 1 2 4v2 =2v 2, (9) where v ms 1 is its speed. For the 6 kg mass the resultant force in the direction of motion is 6g T, so that ( 6g T) 2 = 1 2 6v2 =3v 2, (10) since the 6 kg mass also has speed v ms 1. Adding equations (9) and (10) gives or giving or 12g 8g = 5v 2 4g = 5v 2 v 2 = 4 5 g v 2.8 ms 1. Is it necessary to consider the work done by the tension? Is mechanical energy conserved for this system? If so how can the solution be modified? How would the situation be affected if the pulley was not smooth? What difference would it make if the rope had a significant weight? 144

13 Both this and the falling ball problem illustrate how the 'workenergy' or 'conservation of energy' methods can be used to solve the problem, rather than using Newton's Second Law of motion. This is because, in these examples, you are interested in the speed as a function of distance, and energy methods are quicker in this situation. Activity 5 Kinetic and potential energy Make a list of other activities which involve changes of kinetic and potential energy. If friction or air resistance are included in the model, mechanical energy is no longer conserved, but the work-energy equation is still applicable. h L To illustrate this consider the case of a body moving down a rough inclined plane. An example is a skier coming down a ski slope at speeds when air resistance can be neglected but friction between the skier and the slope is significant. The forces acting on the skier are indicated on the diagram. If the skier starts from rest at the top of the slope and has speed v at the bottom h µn N θ the increase in kinetic energy = 1 2 mv2, mg θ distance down the slope= L = h sin θ. Resolving the forces perpendicular to the plane gives N = mg cosθ and then the law of friction gives F = µ N = µmg cosθ. The work done by the weight and friction = ( mgsin θ) L ( µmg cosθ) L µn N h h = mgsin θ µmg cosθ sin θ sin θ h mgcosθ mgsin θ θ = mgh µmgh cot θ. 145

14 By the work-energy equation or 1 2 mv2 = mgh µmgh cot θ v 2 = 2gh( 1 µ cot θ). Note that v 2 > 0 means 1 > µ cot θ or tan θ > µ, so the angle of the inclination of the plane, θ, must be sufficiently large for the motion to occur. If the surface is smooth then µ = 0 and v 2 = 2gh as before. However, for a rough surface µ 0, so that the speed at the bottom of the slope is less than 2gh which is intuitively obvious. Exercises 6B 1. A man of mass 60 kg climbs a mountain of height 2000 m. What work does he do against gravity? 2. A crane lifts material of mass 500 kg a height 10 m. What is the work done against gravity by the crane? 3. An athlete of mass 80 kg starts from rest and sprints until his speed is 10 ms 1. He then takes off for a high jump and clears the bar with his centre of mass (the point where his weight is assumed to be concentrated) at a height of 2.2 m. How much work has he done up to the moment when he clears the bar? 4. A stone of mass 1 kg is dropped from the top of a building 80 m high. (a) Find the total mechanical energy of the stone before it is released and when it has fallen a distance x metres and its speed is v ms 1. (b) Assuming that energy is conserved show that the speed of the stone v and the distance x are related by the equation v 2 = 20x. (c) Find the speed of the stone when it is halfway to the ground and when it hits the ground. 5. A ball of mass 100 g is thrown vertically upwards from a point 2 m above ground level with a speed of 14 ms 1. (a) With an origin at ground level, find the total mechanical energy of the ball when it is travelling at speed v ms 1 at a height h m. (b) Assuming that mechanical energy is conserved show that v h = 236. (c) Calculate the greatest height reached by the ball. (d) Calculate the speed with which the ball hits the ground. 6.4 Power In some situations it is not only the work done which is important but the time taken to do that work. If you are exercising, the rate at which you exercise may be important. When manufacturers state the time taken for a car to reach a certain speed they are saying something about the rate at which work can be done by the engine. 146

15 The rate at which work is done is called power. If it takes 75 s to move a car a distance of 20 m applying a force of 200 N then the average rate of working or the average power is work done = = 53.3 N ms 1 = 53.3 Js 1. time taken 75 The unit of power is the watt (W) named after the Scottish inventor James Watt, famous amongst other things for harnessing the power of steam. 1 watt is defined as 1 joule per second 1W=1Js 1. The power needed to raise 1 kg vertically 10 cm in 1 s is ( weight) ( distance)= ( 1 g) ( ) 1W (remembering distances are measured in metres) and so the watt is a fairly small unit. For this reason power is frequently measured in kilowatts (kw). Engineers often quote power in horsepower (h.p.) where 1h.p.=746 W. This unit, which is used for cars and lorries, was first introduced by James Watt ( ). It is based on the rate at which large horses, once used by brewers, worked. It is about ten percent greater than the rate at which most horses can work. The French horsepower is slightly less than the British horsepower. Activity 6 Modelling the Log Flume at a Theme Park At the beginning of the Log Flume ride the boat and passengers are carried up the slope by a conveyor belt. In this activity you estimate the power required to get one full boat up the ramp. Here is the data taken on a recent visit. Time taken to raise one boat up the slope = 20 s. (Maximum of 6 persons per boat ) Angle of elevation about 30. Height about 12 m. Mass of boat 150 kg. Taking the average mass of each passenger as 60 kg, calculate the energy required to raise 1 boat and 6 passengers. Hence calculate the power required to raise the full boat. 30 o 12 In some cases when the power is specified, the speed is required at a given time. What is the relationship between power and speed? When pushing the car, in the example at the start of Section 6.4, 147

16 the average power = = ( force ) ( average speed ). If a constant force F moves a body in the direction of the force and at a certain time the body is moving with speed v, then what is the power at that time? Suppose that at time t the distance moved by the body is x metres, then at this time W = work done = F x. Now remembering that power P is rate of doing work P = d dt ( Fx)= F dx dt = Fv,, since F is constant and x varies with t, so P = Fv (11) or Power = Force Speed. Example A car moves along a horizontal road against a resistance of 400 N. What is the greatest speed (in kph) the car can reach if the engine has a maximum power of 16 kw? Solution The car reaches its maximum speed when the engine is at maximum power W. The power needed for the car to move against the resistance at speed v ms 1 is 400 v, using equation (11). The speed v is a maximum when giving = 400 v maximum speed = v = 40 ms 1 = kph 1000 = 144 kph. Activity 7 How fast does the lorry go up the hill? The main frictional resistance F on a lorry is due to the air and depends on the velocity v. For a lorry which has a maximum power of 250 kw, a reasonable assumption is F = 6v 2. What is the speed of the lorry going along a level road when it uses half of its power? If the lorry uses all of its power to go up a hill with slope 1 in 50, 148

17 how do the maximum speeds compare when the lorry is laden to a total mass of 38 tonnes and 44 tonnes? You may find it easiest to solve the equations you obtain for the maximum speeds graphically. Pumping water Often on roadworks pumps are needed to pump water out of the ground. A pump is required to raise 0.1 m 3 of water a second vertically through 12 m and discharge it with velocity 8 ms 1. What is the minimum power rating of pump required? Here the work done is in two parts: the water is raised through 12 m and is also given some kinetic energy. Since 1 m 3 of water weighs 1000 kg, the mass of water raised is 100 kg. The work done in 1 second against this weight of water is (mass raised) g (height raised) = = J, taking g = 10 ms 2. The kinetic energy given to the water in 1 second is = 1 (mass raised) (speed)2 2 = ( 8)2 =3200 J. Hence the total work done in 1 second is = J. The power required is 15.2 kw. If the pump is 100% efficient this is the rate at which the pump is working to raise the water. In practice some power is lost and pumps work at less than 100% efficiency. Here power output efficiency = 100% power input If the pump is only 60% efficient, to raise this water it would need to be working at a rate of = Js 1 = 25. 3kW. Since some power is lost, 15.2 kw is the minimum power rating of the pump. 149

18 Exercise 6C 1. A crane raises a 5000 kg steel girder at 0.4 ms 1. Assuming that work is not lost in driving the crane, at what rate is the crane working? 2. A train of mass kg is travelling at 30 ms 1 up a slope of 1 in 100. The frictional resistance is 50 N per 10 3 kg. Find the rate at which the engine is working. 3. A car of mass 1000 kg moves along a horizontal road against a resistance of 400 N and has maximum power of 16 kw. It pulls a trailer of mass 600 kg which is resisted by a force of 300 N. When the speed is 24 kph and the engine is working at maximum power find the acceleration of the car and trailer and the pull in the tow rope. 4. A lorry of mass 5000 kg with an engine capable of developing 24 kw, has a maximum speed of 80 kph on a level road. Calculate the total resistance in newtons at this speed. Assuming the resistances to be proportional to the square of the speed, calculate in kilowatts the rate of working of the engine when the truck is climbing a gradient of 1 in 200 at 60 kph and has, at that instant, an acceleration of ms Two tugs are pulling a tanker in a harbour at a constant speed of 1.5 knots, as shown. Each cable makes an angle of 12 with the direction of motion of the ship and the tension in the cable is 75 kn. Calculate the rate of working of the two tugs. (1 knot = kph ) o Find the power of a firepump which raises water a distance of 4 m and delivers 0.12 m 3 a minute at a speed of 10 ms A pump is required to raise water from a storage tank 4 metres below ground and discharge it at ground level at 8 ms -1 through a pipe of cross sectional area 0.12 m 2. Find the power of the pump if it is (a) 100% efficient (b) 75% efficient. 6.5 Elasticity Hooke s Law So far strings have been assumed to be inextensible. Whilst for some strings this assumption is reasonable, there are other materials whose lengths change when a force is applied. Activity 8 How do materials behave under loading? You will need wire, coiled springs, shirring elastic, masses and a support stand for this activity. By hanging different masses from the ends of the given materials, investigate the relation between the mass m, and the change in the length x, from the original length. Display your results graphically. Test your specimens to see if they return to their original length when the masses are removed. Keep your data for use in Activity

19 All your specimens are materials which can behave elastically. When you pull the material its length increases but it returns to its original length, sometimes called its natural length, when released. For some of your specimens you may have obtained straight lines such as either (a) or (b), whereas for other specimens a curve like (c) may have occurred. Now Extension = stretched length natural length. If you add too many masses to the wire it either breaks or keeps extending to the floor. This behaviour occurs when you pass the elastic limit and in the case when the wire keeps extending the wire is behaving plastically. When you apply a force to the end of an elastic string and stretch it the string exerts an inward pull or tension T which is equal to the applied force. In case (a) if l is the natural length, L the final length and m is the total mass attached then you will have seen from Activity 8 that the extension x ( = L - l) and m are linearly related by m = qx, (12) where q is the gradient of the straight line. Although the strings and springs used in Activity 8 have masses, they are very much less than those which are added and so can be neglected. When a mass attached to the end of a specimen is in equilibrium two forces act, the tension T upwards and the weight mg downwards. Since the mass is at rest or T mg = 0 T = mg. (13) Eliminating m between equations (12) and (13) gives T = ( qg)x (14) so that T is directly proportional to x. Putting k = qg equation (14) becomes T = kx. (15) This is known as Hooke s Law. The constant of proportionality k is called the stiffness. This is an experimental result which holds for some materials over a limited range, in some cases only for small extensions. Other materials behave elastically for large extensions but the relation is not linear, e.g. curve (c) above. load m load m load m b a extension x c extension x elastic limit extension x T mg 151

20 Putting k = λ l, an alternative form of equation (15) is T = λ l x. (16) Whereas k depends inversely upon l, the constant λ, called the modulus of elasticity, does not depend upon the length but only on the material. Different materials have different values of λ. When x = l, λ = T and so λ is equal to the tension in the string whose length has been doubled. λ has the units of force and so is measured in newtons. The stiffness k = λ l is measured in Nm 1. material modulus of elasticity aluminium copper brass mild steel Some values of the modulus of elasticity for wires of diameter 1 mm made from different materials Activity 9 How stiff are the materials? From your data in Activity 8, where appropriate, determine the values of k and λ for the specimens. Example An elastic string of natural length 60 cm has one end fixed and a 0.6 kg mass attached to the other end. The system hangs vertically in equilibrium. Find the stiffness of the string if the total length of the string in the equilibrium position is 72 cm. Solution The forces acting on the mass are the tension T upwards and its weight mg downwards. Since the mass is at rest T 0.6g = 0 or T = 0.6g = 6 N. T The extension of the string = 12 cm = 0.12 m and so using Hooke s Law 0.6g T = k( 0.12). Equating the expressions for T or 0.12k = 6 k = = 50 Nm

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