Chap 4-5-6 Mechanical Energy and Power PDF

MECHANICAL ENERGY AND POWER 79 CHAPTER 4 MECHANICAL ENERGY AND POWER Niels Wegge © 2017 Box 4.1 Import...

MECHANICAL ENERGY AND POWER 79 CHAPTER 4 MECHANICAL ENERGY AND POWER Niels Wegge © 2017 Box 4.1 Important formulae and numbers to notice Principle of conservation of energy and remember for this chapter: The total energy of a closed system Mechanical energy: 𝐸mech = 𝐸kin + 𝐸pot remains constant in time. Kinetic energy: 𝐸kin = 12𝑚𝑣 2 Potential energy: 𝐸pot = 𝑚𝑔ℎ Principle of conservation Acceleration of gravity: of mechanical energy m World average 𝑔 = 9.8 s2 The sum of kinetic and potential energy Denmark 𝑔DK = 9.82 m s2 of a system that is only 𝐸 Power: 𝑃= under influence of gravity 𝑡 J remains constant in time. SI power unit: W= s kg m2 SI Energy unit: J= s2 Another energy unit: 1 kWh = 3.6 MJ 4.1 Conservation of energy One of the most important principles of physics is the one stated above; that of conservation of energy. In its extreme form it says that no energy is ever lost and no energy will ever go missing: the total number of joules of energy in the Universe is the same now as it was at the time of the Big Bang and as it will be a billion years from now. We cannot prove that this law is universally true as we cannot really wait a So what if the billion years and also cannot measure the energy of a Then the total amount of energy in the Universe, or system does system not remain isn’t of all closed systems in the world. So far, constant? closed! however, no scientist has been able to prove that it is not true: we have never seen a closed system in which energy was not conserved – and therefore we do have faith in the principle of energy conservation. 80 CHAPTER 4 In fact, when the strange neutrino particle was first postulated in 1930 (by Austrian physicist Wolfgang Pauli), this was in a rather desperate attempt to re-establish energy conservation in certain radioactive reactions called beta decays. These decays seemed to violate the principle of energy conservation, as there was measurably more energy present in the system before a beta decay than after. Energy was disappearing without a trace – unless perhaps an extra, but unknown, invisible and undetectable particle sneaked the missing energy out? The dilemma was deep. Either physics would have to give up the principle of energy conservation, or it would have to accept these ghostlike, apparently undetectable, so called neutrinos. But, the philosophers argued, can anything that is undetectable and unmeasurable really be claimed to exist? Fortunately the neutrino was finally directly observed in experiments (but only several years later, in 1956), and the principle of energy conservation survived. It is common to speak about “using energy” or “spending energy” when performing an arduous task. This wording seems to indicate A closed that we end up having less energy than when we started doing the system allows task. But in the big picture we still “have” the same amount of no matter or energy to flow energy as before – the energy has merely changed from one form in or out. to another. Often it changes to a less useful form (such as thermal energy) and therefore in a sense is “lost” because we cannot utilize it any more. However it is important to emphasize that when we look at any closed system (which means anything that does not communicate with its surroundings: no flow of matter or energy in or out), then the total amount of energy in that system remains constant over time. For a simple example, consider a small, hot stone placed in a cup of cold water. The water will get warmer and the stone colder. If the cup is a perfectly insulating thermo cup with the lid on, then the system consisting of cup, water and stone will be closed. Even though the stone is losing energy as it cools, the sum of the thermal energies of the stone and the water and the cup will remain constant. Any energy flowing out of the stone is picked up by the water and the cup. As soon as we take off the lid, the system is not isolated anymore and may receive or lose energy as well as matter. Another example is a car applying the brakes so that it slows down. The kinetic energy (motion energy, see below) of the car decreases, but the energy does not disappear – rather, it is transformed into thermal energy in the metal of the brakes that heat up. In this way, the total energy of the system (the car including its brakes) remains constant. There are several forms of energy. We have already studied thermal energy in detail in Chapter 2, where we also did some calculations with electric energy. Chapter 3 looked at the electromagnetic energy in sunlight and starlight (visible, infrared, microwaves, radio waves, X-rays, gamma radiation) and from other black bodies. Other forms of energy include nuclear energy and chemical energy, but in the present chapter, we will focus entirely on two forms of mechanical energy: kinetic and potential energy. We will begin by looking at energy conservation in the closed system of a basketball in a gym. In the next sections, we then look into kinetic and potential energy in other contexts.  MECHANICAL ENERGY AND POWER 81 4.2 Conservation of mechanical energy Fig. 4.1 shows the trajectory of a basketball that has been tracked on a video in LoggerPro as it flies through the air and bounces on the floor. Each dot represents a frame of the video and gives us a set of coordinates (𝑥 and 𝑦). A plot of 𝑥 and 𝑦 as a function of time can be seen in Fig. 4.2. Figure 4.1 Still from video of basketball shot. Figure 4.2 The 𝑥- and the 𝑦-coordinate The trajectory is shown (Vernier Software). of the basketball as a function of time. Screenshot from LoggerPro. Looking at the plots in Fig. 4.2, we observe the following:  The 𝑦-coordinate (blue) measures the height of the ball over the floor. It increases until time 𝑡  0.65 s; then decreases until 𝑡  1.5 s; then increases again, etc. This is of course because the ball first moves upwards, then falls back down towards the floor, etc. In particular we see that the ball is at maximum height at the time 𝑡  0.65 s and hits the floor at 𝑡  1.5 s.  The 𝑥 -coordinate (red) measures the position of the ball along the length of the gym. It increases all the time. This is because the ball keeps moving to the right (while at the same time going up and down).  The 𝑥-curve consists of two straight lines; one before the ball hits the floor, one after. A straight line in the A straight line in a time-position (𝑡, 𝑥)-graph means that every second the ball travels a graph means fixed number of meters in the 𝑥-direction: the 𝑥-speed constant speed! is constant. The reason why the 𝑥-speed of the ball is constant is that there is no horisontal force significantly affecting the ball while in flight.  The 𝑥-graph becomes less steep after 1.5 s, so the 𝑥-speed becomes less after the bounce: some energy must be “lost” during the bounce itself.  In the plot for the 𝑦-position there are no straight lines, so the up-down speed is continually changing. Changing speed means acceleration – in this case caused by the force of gravity.  The 𝑦-speed starts out decreasing (the gradient of the blue curve gets less and less) until the time 𝑡  0.65 s, which is when the ball turns around and starts falling downwards again. Exactly at this point the 𝑦-speed is zero.  As the ball falls towards the floor the 𝑦-curve gets steeper and steeper, indicating that the ball moves faster and faster. 82 CHAPTER 4 Notice how much we can tell about the speed of the ball by looking at its position graphs! This art will be perfected in Chapter 5. After this qualitative analysis we will now calculate the kinetic, potential and mechanical energies of the ball. The formulae used are 1 𝐸kin = 𝑚𝑣 2 𝐸pot = 𝑚𝑔ℎ 𝐸mech = 𝐸kin + 𝐸pot 2 In the basket ball example, the symbols represent  𝑚: mass of the basketball, 𝑚 = 0.70 kg.  𝑣: speed of the basketball. LoggerPro calculates 𝑣 from the gradients of the position graphs.  𝑔: acceleration of gravity, 𝑔 = 9.8 m s–2 (more about 𝑔 in Chapter 5).  ℎ: height of the ball over the floor, i.e. the 𝑦-coordinate. To see the time variation of 𝐸kin , 𝐸pot and 𝐸mech for the basketball, a plot has been made. Fig. 4.3. shows that  the kinetic energy de- creases when the potential energy increases – and vice versa.  the mechanical energy stays relatively constant at 32 J until the ball hits the ground.  after the bounce, the mechanical energy is again relatively constant, now at a lower level of about 17 J.  at the time of impact, the curves momentarily be- Figure 4.3 Kinetic (red), potential (purple) and total mechanical energy (blue) of the basketball as a function of come very complex, before time. The total mechanical energy stays constant at 32 J stabilizing in a pattern before and at 17 J after the bounce. Screenshot from similar to the one before LoggerPro. the impact.  the graph for kinetic energy never decreases to zero, neither before nor after the bounce. This is because the ball always has a non-zero speed in the 𝑥-direction. The minimum of the 𝐸kin - graph is much closer to zero after the bounce because the 𝑥-speed is much less after the bounce than before. I like decoding The ruggedness of the curves in Fig. 4.3 can be explained by the graphs! non-perfect precision with which the dots were manually placed on the video frames in the first place.  MECHANICAL ENERGY AND POWER 83 To summarize: Fig. 4.3 provides empirical confirmation1 of the principle of conservation Principle of conservation of mechanical energy: of mechanical energy. 𝐸mech of the basketball The sum of kinetic and is constant while it is out of contact with its potential energy of a surroundings (before and after, but not during system that is only under influence of gravity the bounce). The reason that energy remains constant in time. conservation breaks down at around 𝑡 = 1.5 s is that the basketball system stops being isolated at this time: it interacts with the floor. 32−17 During the bounce the mechanical energy goes down from 32 J to 17 J. This is = 0.46875 = 47 %. 32 This large percentage of energy is not “lost” but is converted into thermal energy in the ball (and a little bit in the floor)2 as the ball compresses and reshapes during the bounce: the ball heats up slightly. If we could keep track of this heat as a function of time and plot it in a graph alongside 𝐸kin and 𝐸pot , we would see all three energy graphs add up 32 J both before, during and after the bounce. This would then be an empirical confirmation of the general principle of conservation of energy in this particular case. Here is food for thought: do you think the energy of the basketball drops 47 % in each of the next bounces, too? If so, the heights of the bounces would decrease exponentially, wouldn’t they? You are invited to set up an experiment to investigate this  𝟏 4.3 Kinetic energy: 𝑬𝐤𝐢𝐧 = 𝒎𝒗𝟐 𝟐 What is in a name? In Greek,  means motion (now you understand why a cinema it called a cinema…), so kinetic energy means energy stemming from how a body moves, more than from where 1 it is. But why does the formula for kinetic energy have the form 𝐸kin = 2 𝑚𝑣 2 and not some other form? It seems reasonable that the energy due to motion should depend on mass and speed, but why 1 speed squared, and why the factor 2 in front? If you go on to study physics next year, you will certainly learn more about the reasons for this. To justify the formula in the present course, we will simply look at what happens to the basketball analysis above if we manipulate the formulae. Taking another 1 1 factor, e.g. 1 or 3 instead of 2 to calculate kinetic energy, makes the patterns seen in Fig. 4.3 change radically: 1 Empirical means based on measurements, not on speculation. 2 A minute amount of energy, too, is carried away in the form of sound energy from the “thud” as the ball hits the floor. 84 CHAPTER 4 Figure 4.4 Modification of Fig. 4.3, now with a wrong formula for “kinetic energy”: 1 1 𝑚𝑣 2 (left) and 𝑚𝑣 2 (right) instead of 𝑚𝑣 2. If kinetic energy were really defined 3 2 in this way, the total mechanical energy would no longer be conserved. Fig. 4.4 illustrates that with this alternative (wrong) definition of kinetic energy, the total mechanical energy is not conserved as it was before: the blue curves are not horizontal anymore. A similar pattern would be seen if the power of 𝑣 is changed from 2 to another number. Therefore, apparently, we 1 would lose the principle of energy conservation if the formula for kinetic energy was not 𝐸kin = 2 𝑚𝑣 2. A good reason for using exactly this formula! Example 4.1 Kinetic energy of a car You convert A car of mass 𝑚 = 2500 kg drives at the speed 𝑣 = 50 km h–1. We speeds from km h m want to find the kinetic energy of the car and start by converting to by s the speed to the SI-unit m s–1: dividing by 3.6 km 1000 m 50 m m 𝑣 = 50 = 50 × = = 13.89 = 14 m s −1 h 3600 s 3.6 s s Hence the kinetic energy is 1 1 m 2 kg m2 𝐸kin = 2𝑚𝑣 2 = × 2500 kg × (13.89 ) = 241165 s2 = 240 kJ The beauty of unit 2 s conversion: kg m2 s −2 = J We have used a unit conversion, kg m2 s–2 = J, that you will learn in Strange but true! Chapter 5, and we have rounded to 2 significant figures. If the speed of the car increases by 10 % (from 50 to 55 km/h) the 2 kinetic energy will go up to 12×2500×(3.6 55 ) = 291763 J; an increase of 291763−241165 241165 = 0.2098 = 21 %. This example illustrates why it is considerably more dangerous to crash at an even slightly higher speed. Keep driving slowly!  MECHANICAL ENERGY AND POWER 85 Example 4.2 Relative increase of kinetic energy: scaling The factor 𝑣 2 in the formula for kinetic energy causes the kinetic energy to quadruple if the speed doubles, and if 𝑣 goes up by a factor of 3, then 𝐸kin increases 9-fold. Similarly, if 𝐸kin has decreased by a factor 16, it is because the speed has decreased by a factor 4. Numbers are not always nice and square as above. If, for example the kinetic energy has increased by the factor 6, then it is because the speed has increased by the factor √6 = 2.4 (2 sig.figs.). Percentage increases are easy to deal with, too, and they do not need to involve lengthy calculations as in Example 4.1. If the speed, 𝑣, increases by 10 %, then the new speed is 1.1𝑣, right? And therefore 1 1 the new kinetic energy is 2 𝑚(1.1𝑣)2 = 1.12 × 2 𝑚𝑣 2: you multiply the old one by 1.1 squared. As 1.12 = 1.21, the kinetic energy has gone up by 21 %, same result as in Example 4.1. Example 4.3 Kinetic energy, “backwards” In Example 4.1 we did a straight forward calculation of the kinetic energy, given the speed and the mass. We often run into the converse (or “backwards”) problem of knowing the kinetic energy, and wanting the speed. For example, if a basketball weighing 𝑚 = 0.70 kg is moving with the kinetic energy 𝐸kin = 24 J, then its speed can be found by rearranging the formula: 1 𝐸kin = 2 𝑚𝑣 2 If your kinetic energy is 𝐸 , then your speed 2𝐸kin = 𝑚𝑣 2 is the square 2𝐸kin root of 2𝐸/𝑚. = 𝑣2 𝑚 2𝐸kin 2×24 J 𝑣=√ =√ = 8.281 m s−1 = 8.3 m s −1 (2 s.f.) 𝑚 0.70 kg Notice that 𝐸kin sits under the square root. If the ball had moved at only 8.0 J (which is 3 times less 8.281 m s−1 than 24 J), then the speed would be √3 times less: = 4.8 m s−1. And if 𝐸kin had been 10 % √3 less than 24 J (i.e. 0.90 × 24 J), then the speed would have been √0.90 × 8.281 m s −1 = 7.9 m s −1. 4.4 Potential energy: 𝑬𝐩𝐨𝐭 = 𝒎𝒈𝒉 Kinetic energy is determined by how an object moves – potential energy depends on where an object is. It costs me energy to lift a heavy box from the floor to its storage place on a shelf – and I would get that energy back in a painful form if the box fell down and hit me. The energy I have invested in lifting the box is thus potentially available even after I have deposited the box on the shelf. This potential energy that I have transferred to the box clearly depends on the force of gravity that I am up against, as well as how high I have lifted the box. In section 5.8 we learn that the force of gravity on an object 86 CHAPTER 4 of mass 𝑚 can be calculated by 𝐹grav = 𝑚𝑔, where 𝑔 the acceleration of gravity, and that is why lifting something to the height ℎ adds the potential energy 𝑚𝑔 × ℎ: 𝐸pot = 𝑚𝑔ℎ Looking at the units on the right hand side of the formula, we get kg × m s–2 × m = kg m2 s–2, i.e. joule as in the case of the kinetic energy. Example 4.4 Potato overhead, or “how does one joule feel?” Pick up a potato from the floor and lift it above your head. If the potato weighs 50 grams (0.050 kg) and you lift it to two meters above the floor, then you have increased its potential energy by 𝐸pot = 𝑚𝑔ℎ = 0.050 kg × 9.8 m s–2 × 2.0 m = 0.98 kg m2 s–2 ≈ 1 J This may be a good way of “feeling” what one joule of potential energy is. Try it! Example 4.5 Falling potato If you let go of the 50 g potato in your hand, then it becomes an object only under the influence of gravity (neglecting air resistance) – and we can therefore appeal to the principle of conservation of mechanical energy. If the potato falls from two meters, then its total mechanical energy will remain at 1.0 J during the fall, because it was 1.0 J in the beginning (as we know from the calculation in Example 4.4). As the potato falls, its potential energy decreases, and therefore, to keep the total mechanical energy unchanged, the kinetic energy must increase. This means that the speed increases. Half way down, the potential energy would be half a joule, and therefore the kinetic energy would also be half a joule. Just before the potato hits your foot, its potential energy is 0 J, and therefore the kinetic energy must be 1.0 J. This converts to a speed as in Example 4.3: 2𝐸kin 2×1.0 J 𝑣=√ 𝑚 = √0.050 kg = 6.32 m s−1 = (6.32 × 3.6) km h−1 = 23 km h−1 Ouch! If you did the same experiment on the Moon, where the acceleration of gravity is 6 times less than on the Earth, the initial potential energy of the potato would be 6 times less, and therefore the final speed when it hits your moon boot would be √6 = 2.4 times less. Example 4.6 Falling objects in general – setting up and solving the energy conservation equation Other objects dropped from two meters above the ground, as the potato from Example 4.5, also hit the ground at 6.3 m s–1, regardless of the mass – at least if air resistance is negligible. Indeed, if the mass is 𝑚, and it is dropped from the height ℎ, then its initial mechanical energy is 𝐸mech,1 = 𝐸kin,1 + 𝐸pot,1 = 0 + 𝑚𝑔ℎ = 𝑚𝑔ℎ  MECHANICAL ENERGY AND POWER 87 During the entire fall, the total mechanical energy is conserved and therefore stays constant at the initial value 𝑚𝑔ℎ. Immediately before the object hits the ground, the mechanical energy is 1 1 𝐸mech,2 = 𝐸kin,2 + 𝐸pot,2 = 2 𝑚𝑣 2 + 0 = 2 𝑚𝑣 2 1 and as 𝐸mech,1 = 𝐸mech,2 , we get 𝑚𝑔ℎ = 2 𝑚𝑣 2. The mass conveniently cancels out from both sides of the equation: 1 If you fall from height ℎ 𝑚𝑔ℎ = 2 𝑚𝑣 2 then you hit the ground with speed 𝑣 = √2𝑔ℎ, regardless 1 𝑔ℎ = 2 𝑣 2 of your mass. At least if air resistance is small. 2𝑔ℎ = 𝑣 2 𝑣 = √2𝑔ℎ In particular, if we look at ℎ = 2.0 m, we get 𝑣 = √2𝑔ℎ = √2 × 9.8 × 2.0 = 6.3 m s−1 in agreement with the result in Example 4.5. Example 4.7 Throwing up… how high? Anna throws a stone vertically into the air with initial speed 12.6 m s–1. Reversing the argument of Example 4.6, she can easily calculate how high it reaches before falling back: At maximum height, the speed is zero, so the energy conservation equation 𝐸mech,top = 𝐸mech,initial becomes 21 2 1 𝑚𝑔ℎtop + 2𝑚𝑣top = 𝑚𝑔ℎinitial + 2𝑚𝑣initial 2 1 𝑚𝑔ℎtop + 0 = 0 + 2𝑚𝑣initial 1 2 𝑔ℎtop = 2𝑣initial 2 𝑣initial ℎtop = 2𝑔 which yields ℎtop = 8.1 m when inserting 12.6 m s–1. (Please check the calculations.) You should notice that since 12.6 m s–1 is twice the speed we found in Example 4.6, it is to be expected that the height is now four times the two meters of that example. When the stone falls back towards the ground, its speed increases, until it becomes 12.6 m s–1 immediately before impact, same as the initial speed. Energy conservation! 3 3 Assuming that air resistance is negligible. We will see in Chapter 5.7 what happens if it is not. 88 CHAPTER 4 4.5 Roller coasters In most roller coasters, the car is towed to the highest point and released from there at a speed close to zero. The car races down, faster and faster until it reaches the lowest point, and then goes back up, slowing down as it climbs. Usually, there is no motor in the car, and the system is therefore isolated from the surroundings, except for unavoidable, but hopefully small friction. If we ignore friction, the increase and decrease of speed is governed by mechanical energy conservation: 𝐸kin + 𝐸pot = constant, and the car can go back up to the same height as where it started. With friction, however, every hill along the track must be lower than the previous; otherwise the car cannot make it with the amount of energy available. If there is a loop, the speed must be sufficiently high that the car can make it to the top of the loop (the upside-down-point), but to avoid the car falling off the tracks and the passengers out of the car, the speed actually has to be even higher. This is another interesting story, involving circular motion and the special forces at play there.4 Example 4.8 Roller coaster A The launch for a monster roller coaster is C 42 m above the ground and the next hill is 25 m. A car with four passengers starts at B D the top of the roller coaster (in A) with speed 0. The car has no motor and we will Figure 4.5 Side view of roller coaster. The point A is assume that friction is negligible. What is 42 m above the ground, B and D are at ground the speed at B, C and D? level and C is 25 m above the ground. To calculate the speeds, we will use the principle of conservation of mechanical energy. If the total mass of car and its four passengers is, say,5 𝑚 = 850 kg, then the total mechanical energy of the car is 𝐸mech = 𝐸kin,A + 𝐸pot,A = 0 + 𝑚𝑔ℎ𝐴 = 850 kg × 9.8 m s −2 × 42 m = 350 kJ where we have used the information from A (since this is the only point where we know both speed and height). At all points of the ride, the total mechanical energy will be 350 kJ. In particular, at B, where the potential energy is 0, we get 1 350 kJ = 𝐸kin,B + 𝐸pot,B = 2 𝑚𝑣B2 + 0 1 350000 J = 2 × 850 kg × 𝑣B2 2×350000 J 𝑣B = √ 850 kg = 28.7 m s −1 = (28.7 × 3.6) km h−1 = 103 km h−1 – sweet and fast! 4 Centripetal force. This is on the syllabus for the IB Diploma: something to look forward to! 5 (although this really doesn’t matter: we have already seen that 𝑚 will cancel out of this type of calculation)  MECHANICAL ENERGY AND POWER 89 Similarly, at the point C, the energy conservation equation becomes 350 kJ = 𝐸kin,C + 𝐸pot,C = 12 𝑚𝑣C2 + 𝑚𝑔ℎC 1 350000 J = 2 × 850 J × 𝑣C2 + 850 kg × 9.8 m s −2 × 25 m 𝑣C = 18.3 m s−1 = 66 km h−1 (try the algebra!) In D, the car is at the same height as in B, and therefore the speed will be the same as that in B. If the roller coaster were built on the Moon, we could do all the same calculations except that we would have to use to use 𝑔Moon which is six times less than 𝑔Earth. Hopefully you can see from the equations above – without doing the actual calculations! – that the total mechanical energy would be 6 times less, and the speeds in B, C and D therefore √6 times less than those on the Earth. 4.6 Power 𝐸 Power is the rate of energy conversion, energy flow per time, 𝑃 =. The SI-unit for power, watt, is 𝑡 therefore the same as joules per second: W = J s–1. An electric lightbulb of power rating 60 W converts 60 J of electric energy into 60 J of electromagnetic energy (thermal energy and light) every second. If the bulb is on for 10 seconds, the amount of energy converted is 𝐸 = 𝑃 × 𝑡 = 60 J s–1 × 10 s = 600 J. The concept of power is simple, and we have worked with power already, doing numerous calculations without a formal definition. In Chapter 2, for example, we met electric kettles of a certain power (e.g. 2000 W), and in Chapter 3 we learned that the power of the Sun is about 0.4 billion billion billion W, so that the solar power reaching the atmosphere here on Earth is 1360 W per m2. Example 4.9 Kilowatt-hours If you keep a 1000 W electric oven turned on for one hour, it will have 1 kWh = 3.6 MJ converted the energy 𝐸 = 𝑃 × 𝑡 = 1 kW × 1 h = 1000 W × 3600 s = 3600000 J = 3.6 MJ. This amount of energy is called one kilowatt-hour for obvious reasons. If the 60 W lightbulb is constantly on for a week (you forgot to turn it off when going on vacation…), then the energy “consumption” is 𝐸 = 𝑃 × 𝑡 = 60 W × 7 × 24 h = 10080 Wh = 10 kWh (2 sig.figs.) Electricity bills are usually paid by the kilowatt-hour. You are encouraged to check out how much your electricity company charges you for 1 kWh – and how much your family spends on electricity per year and per day! 90 CHAPTER 4 Example 4.10 Walking and running the stairs To climb a flight of stairs takes some extra energy compared to when walking straight: the potential energy of your body must increase. For instance, if your mass is 60 kg, going up 100 steps of 20 cm height will increase your potential energy by m Δ𝐸 = 𝑚𝑔ℎ = 60 kg × 9.8 s2 × (100 × 0.20 m) = 11760 J = 12 kJ (2 sig.figs.) Doing that slowly is no big deal as 12 kJ is no large amount of energy for a person, but if you do it fast – running up the stairs – you need a lot of power. To walk up those stairs leisurely in 2 minutes takes Δ𝐸 11760 J an extra power output from your muscles of 𝑃 = Δ𝑡 = 120 s = 98 W as compared to walking straight, Δ𝐸 11760 J but running up in half a minute would require an extra power of 𝑃 = Δ𝑡 = 30 s = 392 W. Example 4.11 Power of a crane A crane lifts a box at the speed 1.6 m s–1. The mass of the box is 2300 kg. What is the power output of the crane? Every second the height of the box increases by 1.6 m and thereby its potential energy goes up by 2300 kg × 9.8 m/s2 × 1.6 m = 36064 J = 36 kJ (2 sig.figs.). If the speed is constant, the kinetic energy remains unchanged and the motor of the crane needs to generate 36 kJ every second for potential energy. The Figure 4.6 Crane lifting a box at power output of the crane is therefore 36 kW. constant speed. In the beginning, when the box is accelerated from rest to 1.6 m s–1, the crane also has to supply energy for the increasing kinetic energy, and so the initial power of the crane is larger than 36 kW. Example 4.12 Accelerating car A car accelerates from zero to 100 km/h in 10.0 s. How powerful is the car? If the mass of the car is 1200 kg, the kinetic energy goes from zero to 1 1 100 2 2 𝑚𝑣 2 = 2 × 1200 × ( 3.6 ) = 462963 J = 463 kJ (3 sig. figs) 462.963 kJ during those ten seconds. The corresponding power is = 46.3 kW. In practice, however, the 10.0 s engine must yield a much higher power, as a substantial amount of energy is lost due to air resistance, friction as the tyres roll over the road, friction in the engine and transmission, and heat loss in the combustion. Electric appliances such as headlights, radio and – especially! – the car’s air condition also take their share of power. Car power is traditionally measured in terms of horse power: 1 hp = 735 W. The power 46.3 kW is the 46300 same as 735 = 63 hp. See exercise 233.  MECHANICAL ENERGY AND POWER 91 4.7 Overview The formulae Qualitatively speaking… Mechanical energy: 𝐸mech = 𝐸kin + 𝐸pot Potential energy is proportional to the height. If the height doubles (triples), then Potential energy: 𝐸pot = 𝑚𝑔ℎ 1 the potential energy doubles (triples). Kinetic energy: 𝐸kin = 2 𝑚𝑣 2 Kinetic energy is proportional to the speed squared. If the speed doubles (triples) The principles [quadruples], then the kinetic energy becomes four (nine) [sixteen] times larger. A mechanical system may lose energy through friction and gain energy from a motor or through the work of some Fa l l external force. But if the system is closed – only under influence of gravity – then the If there is no air resistance, then total mechanical energy remains constant the speed of impact depends on i all the time: the height from where you fall, not on your mass: n 𝐸mech = constant 𝑣 = √2𝑔ℎ. If all forms of energy are included, then the total energy of any closed system is con- g This formula is a consequence of served: mechanical energy conservation. 𝐸total = constant … I believe in Power and energy energy conservation, Δ𝐸 yes I do!  𝑃= Δ𝑡  Δ𝐸 = 𝑃 × Δ𝑡 In particular, one kilowatt-hour is 1 kWh = 1000 W × 1 h = 3600000 J = 3.6 MJ Your power is the I am powerful when energy you spend I can convert a lot per time. of energy fast! 92 CHAPTER 4 4.8 Exercises 1. Straightforwardly kinetic Calculate the kinetic energy of a … a. … car with mass 2200 kg and moving at 10 m s–1. [110 kJ] b. … car with mass 2200 kg and moving at 20 m s–1. [440 kJ] c. … car with mass 2200 kg and moving at 50 km h–1. [210 kJ] d. … bullet with mass 4.0 g and travelling at 200 m s–1. [80 J] 1. Backwards kinetic a. A tennis ball weighs 0.059 kg. The fastest tennis serve ever was when Australian Samuel Groth sent off the ball with kinetic energy 𝐸kin = 160 J in 2012. Calculate the speed of the ball. [265 km/h!] Make reasonable assumptions about mass and then calculate the speed of … b. … a 1.5 L soda bottle (full) hitting the ground with kinetic energy 𝐸kin = 15 J. c. … a classmate running with a kinetic energy of 1.0 kJ. 2. Straightforwardly potential Calculate the increase in potential energy of … a. … a 5.0 kg box that you lift from a table to a shelf 1.7 m higher up. [83 J] b. … a 250 g chocolate bar that you carried to the top platform of the Eiffel Tower. c. … yourself when you have climbed a table in the classroom. 3. Backwards potential a. You have 100 J available to lift a box weighing 2.0 kg. How high can you get it? [5.1 m] b. If you were on the Moon, where 𝑔 = 1.62 m s–2, how high could you then get it? [31 m] c. On Mars? 5. Powerful machine A machine in a factory yields the power 1500 W. The machine is used for lifting boxes (each weighing 200 kg) from the floor to a shelf 3.0 m above the floor. a. How much energy can the machine yield per minute? [90 kJ] b. How many boxes can it lift per minute? 8. Close inspection of a bounce Look closely at Fig. 4.3. For how long is the basketball in contact with the floor during the bounce? Hint: there are 30 data points per second. [0.17 s] 13. Roller coaster in general a. Do the calculations of Example 4.8 with general mass 𝑚 instead of the concrete mass 850 kg. (You should be getting the same results!) b. Which number do you think the next exercise will have?  MECHANICAL ENERGY AND POWER 93 21. Rain and hail A raindrop falls from a cloud 500 m up. a. Use the formula 𝑣 = √2𝑔ℎ to calculate the speed of the raindrop immediately before it hits the ground (or you), assuming no air resistance. Give the answer in km h–1. [360 km h–1] b. Obviously, this speed is completely unrealistic. Estimate the speed of raindrops from your own experience and explain why the formula above cannot be used. c. Optional: read Chapter 5.7 about air resistance and terminal speed.  34. How high does it get? a. A ball of mass 0.15 kg is fired vertically into the air. Its initial speed is 19 m s–1. Calculate the initial kinetic energy of the ball. [27 J] b. Assuming air resistance is negligible, write down the value of the total mechanical energy of the ball at all times during its flight. c. Hence calculate the maximum height reached by the ball. [18 m] d. With what speed does the ball hit the ground? e. How would air resistance affect each of your answers to the above questions? 55. Dropping bottles (don’t!) An empty bottle falls from the iconic tower Rundetårn in Copenhagen.  Height of tower: 35 m  Initial speed of bottle: zero  Mass of bottle: 0.100 kg  Acceleration of gravity in Denmark: g = 9.82 m s–2 a. Fill in the blanks below, rounding off to 3 significant figures. b. At which speed (in km h–1) will the bottle hit the ground (or a pedestrian)? c. Discuss what would happen if you had dropped a ping pong ball instead of a bottle. Height of bottle Potential energy Kinetic energy of Total mechanical Speed of the over the ground of the bottle the bottle energy of the bottle bottle ℎ/m 𝐸 pot / J 𝐸 kin / J 𝐸 tot / J 𝑣 / (m s–1) 35.0 0 30.0 20.0 20.0 25.0 0 94 CHAPTER 4 89. Basketball toss A basketball is tossed vertically into the air as shown in the video clip below (Vernier Software). The position of the ball is tracked with a motion sensor which gives the following graphs of the vertical position of the ball over the table and its speed, both as a function of time. a. Use the velocity graph to determine the duration of the time the ball spends in the air without being touched by the girl. [between 0.6 s and 0.7 s] b. Can you see that the velocity of the ball is positive before and negative after 𝑡 = 1.0 s? Explain why this was to be expected from the shape of the position graph. c. The mass of the ball is 0.80 kg. Use one of the graphs to calculate the maximum kinetic energy of the ball. [3.6 J] d. Use the other graph to calculate the maximum potential energy of the ball, relative to the level of the table. [7.8 J] e. State the value of the kinetic energy at the time when the potential energy is maximal. [] f. State the constant value of the total mechanical energy of the ball after it has left the hands of the girl and until she catches it again. g. The graphs of the kinetic, potential and total mechanical energies as a function of time are shown in the diagram below. Compare your answers to questions e and f to the values you can read off the diagram.  MECHANICAL ENERGY AND POWER 95 144. Over the hill – or not… There is a speed limit of 70 km h–1 on the road Kongevejen in Holte, Denmark, as you can see on the photo. On this road, a hill called Geels Bakke has a height of 20 m (dreaded by cyclists). If you drive your car at 70 km h–1, can you then switch off the engine at the foot of the hill and simply roll over Geels Bakke? Justify your answer through calculations of kinetic and potential energies, assuming that the mass of the car is 𝑚. 233. Horse power historically James Watt, perfector of the steam engine in the second half of the 18th century, estimated that a standard horse could turn a horizontal mill wheel 144 times in an hour, thereby raising a weight of 180 pounds in a mineshaft. The weight was suspended from a rope that got wound onto the circumference of the wheel as it turned. a. If the mill wheel was 24 feet of diameter, calculate how many feet the weight was raised per second (yes, it will be useful to calculate the circumference of the wheel…) [3.0 feet s–1] b. Hence estimate the power of the horse. You will need the following conversions:  1 pound = 0.454 kg  1 foot = 0.305 m c. Compare the value you got in question a. to the current standard Hippo value 1 hp = 735 W. It shouldn’t be very different. power? d. Estimate your own maximum power, e.g. by running up the stairs as fast as you can. How many horse powers can you deliver? e. Find out what horse power is called in German and other languages! 377. Bouncy ball Oscar stands on a chair on the pavement and drops a small rubber ball from the height 2.5 m. The mass of the ball is 50 g and its initial speed 0 m s–1. a. Calculate the speed of the ball as it hits the ground. [7.0 m s–1] b. The elasticity of the ball is 85 % – meaning that 15 % of the mechanical energy of the ball is “lost” during the impact. Calculate the kinetic energy of the ball as it bounces back up, immediately after the impact. c. Calculate the maximum height reached by the ball after the impact. [2.1 m] d. How high does the ball go in each of the subsequent bounces? e. According to the general principle of energy conservation, no energy is ever lost. Where, then, does the 15 % of the mechanical energy that is “lost” in each bounce go? f. Challenge: Oscar now actively throws down the ball with an initial speed of 6.0 m s–1. It bounces back up and goes above the initial height of 2.5 m. How much above? 96 CHAPTER 4 610. Baseballing A baseball player hits the ball well and sends it off at an amazing speed of 40 m s–1 (144 km h–1!). The angle with horizontal is 45o and therefore the vertical speed of the ball is 𝑣𝑦 = 40 m s −1 × sin(45°) = 28.284 m s −1 = 28 m s−1 (2 significant figures). The mass of the ball is 𝑚, and we will (unrealistically) assume that there is no air resistance. a. Set up the equation that expresses conservation of mechanical energy in the vertical direction. b. Hence calculate the maximum height reached by the ball. [41 m] o c. Same as b, but now the angle is 30. [20 m] d. How would air resistance affect your result in b and c? 987. Bungy jump When you bungy jump (if you do), your speed increases at a constant rate until the rubber rope starts pulling at you. Your potential energy keeps decreasing until you reach the maximum stretch of the rubber rope. As the rubber rope stretches, elastic energy gets stored in the rope.  Sketch rough – but realistic – graphs of kinetic, potential, elastic energy as function of time. You can assume that the first two (endless!) seconds is a free fall before the rubber rope kicks in. The maximum stretch happens at, say, 𝑡 = 5.0 s. F17. Climbing the stairs a. Siegfried weighs 72 kg and his body can deliver a maximum extra power of 800 W over a short period of time. How quickly can he run up 50 steps? You can assume the height of each step is 25 cm. [11 s] b. Siegfried puts on a backpack that weighs 12 kg. How fast can he now run up the stairs? [13 s] F18. Accelerating car A VW Passat Variant 1.8 Turbo has, according to the manufacturer, the following data:  Length: 4.60 m  Mass: 1375 kg  Max acceleration: 0 – 100 km h–1 in 9.4 s  Max engine power: 110 kW a. Calculate the increase of kinetic energy and hence the power required to give the max acceleration listed. b. Why is this power less than the listed max engine power? c. Convert the listed engine power from kilowatt to horsepower.  MECHANICAL ENERGY AND POWER 97 F19. Revision / understanding a. Write down the formulae for kinetic, potential and mechanical energy. b. Calculate the kinetic energy of an object of mass 2.0 kg and moving at 10 m s–1. c. Is it correct that your kinetic energy quadruples if your speed doubles? d. What happens to your kinetic energy if you speed gets three times higher? e. Sketch a graph of kinetic energy as a function of speed and use it to argue why it may be a good idea to enforce slightly lower speed limits. Both think about the energy released in a car crash and the issue with breaking distance. f. How much energy does it roughly take to lift a 10 kg box from the floor to a table? g. State the principle of conservation of mechanical energy. h. You throw your phone vertically into the air with the speed 10 m s–1. What will the speed be when you catch it again? i. Use the principle of conservation of mechanical energy to derive the formula 𝑣 = √2𝑔ℎ for the speed of an object that has fallen from height ℎ, starting at speed zero. j. A certain battery delivers 15 J of energy in the time 10 s. Calculate the power of battery. k. A toaster with power rating 1000 W is on for two minutes. How many joules of electric energy does that take? l. A common misconception is that the unit kilowatt-hour (kWh) is a unit for power – not for energy. Explain why this is wrong. And explain why 1 kWh = 3.6 MJ. m. You run uphill at constant speed. Explain why it requires more power if the gradient gets steeper. 98 CHAPTER 5 CHAPTER 5 MOTION Niels Wegge © 2017 Kinematics deals with motion (µ…) and we have already had a lot to say about objects in motion in Chapter 4. There the governing principle was that of mechanical energy conservation, which allowed us to calculate how fast a roller coaster car would go at different heights of its path, how high a ball would get when bouncing back from the floor, etc. We saw how the principle of energy conservation became an equation that we could solve to find the correlation 𝑣 = √2𝑔ℎ between height and speed. In the present chapter, we shall examine motion in closer detail, now as a function of time. What is the Box 5.1 Important formulae and units average speed of the train from Copenhagen to to notice and remember for this chapter Paris? Where will the ball be 1.5 seconds after we throw it? How fast must we reduce speed in order to Motion at constant velocity: avoid colliding with the goat 30 meters down the  No acceleration! 𝑥 road? How long does it take the rocket to accelerate  𝑣= 𝑡 to 100 m s–1 ? When and how should we launch that  𝑥 = 𝑣𝑡 + 𝑥0 rocket, so that the Mars lander it carries actually  (𝑡, 𝑥)-graph is a straight line arrives on Mars instead of getting lost in space? Motion at constant acceleration: 𝑣 We will develop a set of formulae that allow us to  𝑎= 𝑡 calculate answers to questions like the ones above.  𝑣 = 𝑎𝑡 + 𝑣0 This set of formulae provides a model for a moving 1  𝑥 = 𝑎𝑡 2 + 𝑣0 𝑡 + 𝑥0 2 object, from which we can understand how it has  (𝑡, 𝑣)-graph is a straight line moved, and predict how it will move. In particular,  (𝑡, 𝑥)-graph is a parabola we will draw three motion graphs for the moving object: one for the position, one for the speed and SI units: one for the acceleration; all as a function of time.  Position: m Graphical analysis becomes a powerful tool to  Speed: m s–1 answer motion questions. Interestingly, it turns out  Acceleration: m s–2 that each one of those graphs contains sufficient information to recreate the other two.1 Read on… 1 We are at the verge of differential and integral calculus here!  MOTION 99 It proves mathematically convenient to divide motion into three types: One two  Type 1: Motion at constant velocity (zero acceleration): section 5.3. three  Type 2: Motion at constant, non-zero acceleration: section 5.5.  Type 3: Motion at non-constant acceleration: section 5.6 and 5.7. The equations for type 1 and type 2 are already listed in box 5.1 above, and they will be discussed below. Type 3 we can only treat qualitatively in this course – the main example will be that of air resistance and terminal speed, see the optional section 5.7. To get started, we must introduce and carefully define the kinematic variables. 5.1 Displacement, average speed and average velocity Where is the ball right now, where is it going – and how t = 2.5 s fast? To answer the first of these questions we could simply take a snapshot of the ball (ah, there it is!); but it would not help us answer the other two.2 An extra snapshot, however, taken a little later, will reveal how far and in which direction the ball has moved. Fig. 5.1 shows two such snapshots of a ball rolling on a floor. First the ball is at point A(2,1) and t=0s then, 2.5 seconds later, it is at B(5,5). The numbers on the axes are meters, as measured from the corner of a room. Figure 5.1 Two snapshots of a ball rolling on a floor (top view). The We see that the ball has moved 5–2 = 3 meters parallel to axes show the position of the ball one wall of the room and at the same time 5–1 = 4 meters (in meters) relative to one corner of parallel to the other wall. In total, Pythagoras says, it has the room. There is 5 m from A to B. changed its position by √32 + 42 = 5 m. If this took 2.5 s, then we would think that the speed was 5/2.5 = 2 meters per second. And it would be, if the ball moved from A to B in a straight line without changing speed on the way. But we really do not know what happened during the 2.5 seconds between the two snapshots. What if the ball moved along the path shown in Fig. 5.2? Then the ball would certainly have travelled much longer than 5 m, and consequently its average speed would have been much larger than 2 m s–1. Figure 5.2 Actual path of the ball To resolve this problem we need a precise definition of between the snapshots: much “speed” – in fact we need to distinguish between speed and longer than 5 m. velocity, as shown below. 2 Well, if the shutter time is not too short, we might be able to say something about the speed of the ball from the blurriness of that one, single picture… 100 CHAPTER 5 To keep things simple, in this chapter we will mostly look at motion in one dimension, back and forth along a line. A proper treatment of motion in two and three dimensions requires vectors, which you will learn in later mathematics and physics courses. Something to look forward to! Position 1 Position 2 Position / m Figure 5.3 An object moving in a straight line from 𝑥 1 = –1 m to 𝑥 2 = 3 m. The displacement is ∆𝑥 = 𝑥2 − 𝑥1 = 4 m. The hierarchy of kinematic variables:  Position: where the object is, measured on an axis along the direction of motion. o Often denoted by 𝑥 and measured in meters. o When the position is not stationary (constant), it is interesting to graph 𝑥 as a function of the time 𝑡 in a so-called (𝑡, 𝑥) -diagram. This graph illustrates how the position changes with time.  Displacement: the distance between two positions. o Often denoted 𝑥 and measured in meters. o In Fig. 5.3 above, the object has been displaced 4 m from position 1 to position 2: 𝑥 = 𝑥2 − 𝑥1 = 3 m – (–1 m) = 4 m. o Displacement may be positive (position 2 is after position 1), negative (position 2 is before position 1) or zero (position 1 and 2 is the same place). o If the motion is along a curved line, then the displacement is split into components: one for the 𝑥-direction and one for the 𝑦-direction (and one for the 𝑧-direction if in three dimensions).  Distance travelled: the length of the path taken by the object from one position to another. o Often denoted by 𝑠 or 𝑑, always a positive number. o If the object changes direction of motion, then the distance travelled will be longer than the displacement, as in Fig. 5.2. o A moving object that starts and ends in the same position will have traveled a certain distance 𝑠, but its displacement ∆𝑥 will be zero. Distance travelled is always larger than or equal to displacement.  Average speed: the distance travelled divided by the time it took. o Usually denoted 𝑣 (not 𝑠, please!) 𝑠 o 𝑣 = 𝑡 o Average speed is always a positive number.  Average velocity: the displacement divided by the time it took. 𝑥 o 𝑣= 𝑡 (denoted 𝑣, just like speed) o Average velocity may be positive (when 𝑥 > 0), negative (when 𝑥 < 0) or zero (when 𝑥 = 0). o If the object is not changing direction of motion during the time 𝑡, then the average speed is the same as the numeric value of the average velocity. Otherwise, average speed > |average velocity|.  MOTION 101 Example 5.1 Olympic swimmer American swimmer Dana Vollmer set the current world record for women’s 100 m butterfly at the London Olympics in 2012. Her time was an amazing 55.98 s. The swimming pool was 50 m long, and so Dana Vollmer ended up where she started after having completed the 100 m. From a kinematics point of view:  Distance travelled: 𝑠 = 100.0 m 𝑠 100.0 m  Average speed: 𝑣 = ∆𝑡 = 55.98 s = 1.786 m s −1  Displacement: 𝑥 = 0 m ∆𝑥 0m  Average velocity: 𝑣 = ∆𝑡 = 55.98 s = 0 m s −1 Example 5.2 Forgetful physics student Oscar’s way to school is a straight road. His house is at 𝑥1 = 0 m, and his school at 𝑥2 = 2500 m. Oscar rides his bike, and one nice, brainless morning he gets halfway to school before realizing that he forgot his physics book. He turns around and rides home to get the book, then goes straight to school. He is late, but fortunately his first lesson is chemistry, not physics. The whole trip took 15 minutes. What was Oscar’s average speed and average velocity coming to school that morning? school 0 1250 2500 𝑥/m Figure 5.4 Oscar turns around halfway to school to get his physics book.  Distance travelled: 𝑠 = 1250 m + 1250 m + 2500 m = 5000 m  Displacement: 𝑥 = 2500 m  Time taken: 𝑡 = 15 × 60 s = 900 s 𝑠 5000 m  Average speed: 𝑣 = 𝑡 = 900 s = 5.56 m s−1 = (5.56 × 3.6) km h−1 = 20 km h−1 𝑥 2500 m  Average velocity: 𝑣= 𝑡 = 900 s = ⋯ = 10 km h−1 Example 5.3 Speed of the Earth in orbit around the Sun Every year, the Earth completes yet another full trip around the Sun. The orbit is roughly circular with radius 𝑅 = 150 million km. Kinematic analysis (although the motion is not in a straight line):  Distance travelled yearly: 𝑠 = 2𝜋𝑅 = 2𝜋 × 1.5 × 1011 m = 9.425 × 1011 m = 9.4 × 1011 m  Displacement yearly: 𝑥 = 0 m and 𝑦 = 0 m  Time taken: 𝑡 = 365.25 × 24 × 60 × 60 s = 3.156 × 107 s 𝑠 9.425×1011 m  Average speed: 𝑣 = 𝑡 = 3.156×107 s = 29900 m s−1 = 30 km s −1 𝑥 𝑦  Average velocity: 𝑣𝑥 = = 0 m s−1 and 𝑣𝑦 = = 0 m s−1 𝑡 𝑡 102 CHAPTER 5 So there you are, sitting motionless on your chair – moving 30 The Earth travels kilometers through space every second! You also have a speed due 30 km every second. But its to the rotation of the Earth. Try guessing how many kilometers per annual average second that is – and do exercises 17 and 18. velocity is zero. Example 5.4 Average speed A train travels 360 km from A to B in three hours and then 140 km from B to C in two hours. What are the average speeds? 360 km km  From A to B: 𝑣= 3.0 h = 120 h 140 km km  From B to C: 𝑣= 2.0 h = 70 h (360+140) km 500 km km  From A to C: 𝑣= (3.0+2.0) h = 5.0 h = 100 h Notice that the overall average speed, 100 km h–1, is not the average of 120 km h–1 and 70 km h–1! Example 5.5 How to average averages The train from Example 5.4 continues on from C to D, travelling 450 km at average speed 150 km h–1, and then 200 km from D to E at average speed of 100 km h–1. What is the overall average speed of the train from C to E? The distance travelled for the full trip from C to E is 450 km + 200 km = 650 km, and we need to find out how long that takes. Let us determine the times for each of the two sub-trips (C to D and D to E) 𝑠 by making 𝑡 the subject in 𝑣 = ∆𝑡 : Three formulae in one: 𝑠 450 km  From C to D: 𝑡 = 𝑣 = 150 km h−1 = 3.0 h  𝑣= 𝑠 𝑡  From D to E: 𝑡 = = 𝑠 200 km = 2.0 h  𝑠 = 𝑣 × 𝑡 𝑣 100 km h−1  𝑡 = 𝑠 𝑣 Hence, the total trip takes five hours, and the average 650 km speed is therefore 𝑣 = = 130 km h–1. We see, 5.0 h once more, that the average speed is not the average of the speeds! Position, 𝑥 / km Example 5.6 Position graphs Y -train x = 100 km Fig. 5.5 shows the (𝑡, 𝑥)-graphs for two different trains travelling 100 km from X to Y. As you can see, they both -train do the trip in one hour, so they both go at average speed 100 km h–1. During the first half hour the -train (red X t = 1h curve) covers more distance than the -train, and so it must have been moving faster. During the second half Time, 𝑡 / h hour, it is the -train (green curve) that goes faster. Figure 5.5 (𝑡, 𝑥)-graphs for two trains.  MOTION 103 5.2 Instantaneous velocity If we are logging the position of a train as it goes, we can draw a position graph like the one in Example 5.6. Such a (𝑡, 𝑥)-graph of course shows where the train is at any time, but, importantly, it also carries information about the velocity:  If the train moves fast, then its position increases a lot in a short time, and so the (𝑡, 𝑥)-graph will be steep.  Similarly, if the train moves more slowly, the (𝑡, 𝑥)-graph will be less steep.  If the train stops, and doesn’t move for some time, then the position is constant during that time. The (𝑡, 𝑥)-graph will be horizontal.  If the train goes backwards, its position decreases. The graph will have a negative gradient. We can turn this qualitative reasoning into an important quantitative result by recalling that the steepness of a straight line is calculated as the “rise-over-run”: this is how you find the gradient in mathematics. Clearly, on the (𝑡, 𝑥)-graph, the “rise”, ∆𝑥, would be the displacement of the train (or whatever object it describes), and the “run”, ∆𝑡, would be the time taken for that displacement. Voila: rise ∆𝑥 on the (𝑡, 𝑥)-graph, the rise over run is nothing but the average velocity: run = ∆𝑡 = 𝑣. Position, 𝑥 Position, 𝑥 𝑥 𝑥 𝑥 𝑣 = 𝑣= 𝑡 𝑡 x t t Time, 𝑡 Time, 𝑡 Figure 5.6 The average velocity between two positions is the gradient of the straight line joining the two points on the (𝑡, 𝑥)-graph. Fig. 5.6 demonstrates how the average Position, 𝑥 velocity is the gradient of the straight line joining two points, and how, when  𝑡 is small, this is not very different from the 𝑥 𝑣= 𝑡 steepness of the curve itself. We therefore define the instantaneous velocity of a 𝑥 moving object to be the gradient of the 𝑡 tangent of its (𝑡, 𝑥)-graph as shown on Fig. 5.7. Time, 𝑡 When you learn to differentiate in Figure 5.7 Instantaneous velocity at a given time is mathematics, it will be easy for you to find the gradient of the tangent to the (𝑡, 𝑥)-graph. 104 CHAPTER 5 the gradient of the tangent, and thereby Instantaneous velocity is the gradient of the tangent directly calculate the instantaneous velocity. to the (𝑡, 𝑥)-graph! In this PreDiploma course, however, we shall only use the graphical approach to instan- taneous velocities. I wish I could differentiate… Example 5.7 Reading off velocities from the steepness of the position graph The position of a moving particle has been Position 𝑥 / m continuously logged for three seconds as shown in Fig. 5.8. We see that the position increases at a decreasing rate until 𝑡 = 1.5 s. Then the particle turns around and comes back to its original position at 𝑥 = 0 m, arriving at 𝑡 = 3 s. The tangents to the (𝑡, 𝑥)-graph have been drawn at A, B and C (dotted lines). You are invited to check that the gradients of the tangents are 15, 5 and 0, respectively. Oh – gradients in physics usually have units: in this m case the unit is s , in fine agreement with the Time 𝑡 / s fact that gradients of a (𝑡, 𝑥) -graph are Figure 5.8 The position of a particle moving out and velocities. So the instantaneous velocity of m m back along a straight line for three seconds. the particle is 15 s initially (at A), 5 s at B, Tangents are drawn at A, B and C. m The points D and E are symmetric to B and A. and 0 s at the graph’s turning point C. Clearly, as the graph is symmetrical, the gradients at D and E must be –5 and –15 meters per second, respectively. Negative velocities means moving “backwards”. This could be a ball thrown vertically into m the air with an initial speed of 15 s , cf. exercise 16. 5.3 Motion at constant velocity Let us justify the formula 𝑥 = 𝑣𝑡 + 𝑥0. When a car travels at constant speed, say, 10 m s–1, its position increases by 10 m every second. Consequently, its (𝑡, 𝑥)-graph is a straight line with a “rise” of 10 m for every 1 s you “run” out the 𝑡-axis. After, say, 5 seconds, the position of the car will have increased by 510 = 50 m. If the car starts at the position 𝑥 = 100 m at time 𝑡 = 0 s (the initial position), then after 5 seconds the car will be at m 𝑥 = (10 s ) × (5 s) + 100 m = 150 m  MOTION 105 More generally, at time 𝑡 the position will be m 𝑥 = (10 s ) × 𝑡 + 100 m m Writing 𝑣 instead of 10 s and 𝑥0 instead of 100 m, we have the fully generalized formula for the position of an object that moves at constant speed 𝑣 and starts at initial position 𝑥 0: 𝑥 = 𝑣 × 𝑡 + 𝑥0 In a (𝑡, 𝑥)-diagram this is nothing but the formula of a straight line with gradient 𝑣 and 𝑦-intercept 𝑥0. Notice that if two objects move at the same speed, then their (𝑡, 𝑥)-graphs will have the same gradient and therefore be parallel (see Fig. 5.9). Position, 𝑥 / m Motion at constant Car A speed: That’s the 100 m 𝑥 = 𝑣 𝑡 + 𝑥0 equation for a straight line in a (𝑡, 𝑥)- 100 m diagram! Car B Time, 𝑡 / s Figure 5.9 Car A and B drive at the same, constant speed. The initial position of car A and B is 100 m and zero, respectively. Car A is constantly 100 m ahead of car B. Example 5.8 Catching up with the slow car Fig. 5.10 shows the (𝑡, 𝑥)-graphs for two cars. Car Position, 𝑥 / m C drives at constant velocity: its graph is a straight line. The velocity is ∆𝑥 (600−0) m 𝑣C = = (60−0) s = 10 m s −1 ∆𝑡 P 𝑥1 Car D also drives at constant velocity, but starts one minute later than C. If 𝑣D = 15 m s−1 , let us 600 calculate when (𝑡1 ) and where (𝑥1 ) D catches up with C. The cars take the time 𝑡1 and 𝑡1 − 60 s, 60 𝑡1 Time, 𝑡 / s respectively, to travel the distance 𝑥1. Hence: Figure 5.10 Car C gets a head start of 600 m,  Car C: 𝑥1 = 𝑣𝐶 × 𝑡1 but car D catches up with car C at time 𝑡1.  Car D: 𝑥1 = 𝑣𝐷 × (𝑡1 − 60 s) which combine to give 𝑣𝐶 × 𝑡1 = 𝑣𝐷 × (𝑡1 − 60 s). Solving for 𝑡1 gives 𝑡1 = 180 s, and thence 𝑥1 = 10 m s−1 × 180 s = 1800 m. Also see exercises 19 and 20. 106 CHAPTER 5 5.4 Acceleration For every physics variable, it will be interesting to know the rate of change of that variable. We have just seen how the rate of change of the position of a moving object gives us its velocity. Other examples might be mass and temperature. If I weigh 120 kg and am on a diet, I should like to know how fast – at which rate – I am losing weight. How many kg per week? And if I am boiling water in an electric kettle, I would like to know at which rate the temperature goes up. Fast or slow, how many degrees per second? In all cases one would plot the time graph of the variable (time along the first axis) and look for the gradient of the graph. The variable for which we will now study the rate of change is velocity: how fast does the velocity of m an object change? This rate of change we call acceleration, and its unit of course becomes s per m⁄ m s second, i.e. , which is the same as or m s −2. If, for instance, the acceleration of a falling ball is s s2 m m 9.8 s2 , then the velocity of the ball increases by 9.8 s every second. Acceleration is a measure of how much the speed changes – and how fast it happens. If a car accelerates from rest to 100 km h–1 in 10 s, then that acceleration is larger than if it took 20 s. Similarly, if the car accelerates for 5 s, then the acceleration will be larger if the velocity during those 5 s increases by 10 km h–1 than if it only increases by 8 km h–1. Just as we must distinguish between average and instantaneous velocity, we also distinguish between average and instantaneous acceleration: Velocity , 𝑣 Velocity, 𝑣 𝑣 𝑣 𝑎= 𝑡 𝑣 𝑎 = 𝑡 𝑣 t 𝑡 Time, 𝑡 Time, 𝑡 Figure 5.11 The average acceleration from one Figure 5.12 Instantaneous acceleration at a given velocity to another is the gradient of the straight time is the gradient of the tangent to the (𝑡, 𝑣)- line joining the two points on the (𝑡, 𝑣)-graph. graph. Average If the time interval (∆𝑡) during which the velocity acceleration So the SI 𝑣 unit is changes (by  𝑣 ) is sufficiently short, then the is 𝑎= 𝑡 ms–2 ∆𝑣 average acceleration (𝑎 = ∆𝑡 ) is practically equal to the instantaneous acceleration, which is the gradient of the tangent to the (𝑡, 𝑣)-graph.  MOTION 107 Example 5.9 Accelerating car Two cars, A and B, drive down the road at 20 m s–1. Then they enter the freeway and both accelerate to 30 m s–1. Car A does it in five seconds, car B in ten seconds. The velocity increase is the same for both cars: 𝑣 = (30 – 20) m s–1 = 10 m s–1, but the acceleration is larger for car A than for car B: 𝑣 10 m s−1 Car A: 𝑎A = 𝑡 = 5s = 2 m s−2 𝑣 10m s−1 Car B: 𝑎B = = = 1 m s−2 𝑡 10 s m These are average accelerations, as we do not know in detail how the velocity changed from 20 s to m 30 s. Example 5.10 Maximum acceleration – fast cars Most cars can go very fast, but when we talk about “a fast car”, we usually do not think about its maximum speed – we mean one that can accelerate fast. If a car can go from zero to 100 km h–1 in, say, 5.0 seconds (that is very fast!), then its average acceleration is 𝑣 100 km h−1 100 3.6 m s −1 𝑎= = = = 5.5556 m s −2 = 5.6 m s −2 (2 sig. figs. ) 𝑡 5.0 s 5.0 s Next, let us calculate how long it takes a car to go from zero to 100 km h–1 at the even higher 𝑣 acceleration of 10 m s–2. As we want 𝑡, we begin by rearranging the equation 𝑎 = : 𝑡 𝑣 100 km h−1 100 m s−1 𝑡 = = = 3.6 = 2.7778 s = 2.8 s (2 sig. figs. ) 𝑎 10 m s −2 10 m s−2 How far does the car go during those 2.8 seconds? We cannot use the formula 𝑥 = 𝑣 × 𝑡 , because now the velocity is changing all the time. The correct formula for distance travelled at constant acceleration is provided in section 5.5 – read on. How fast can a car accelerate? To increase the speed fast, the kinetic energy has to increase fast, and for that you need a powerful engine. Once you get over a certain acceleration, power, however, the maximum acceleration is not limited by engine power but by the traction between the rubber tires Figure 5.13 Propelled by jet engines, the and the road: if the wheels just spin, you do not get Thrust Supersonic Car holds the record for much acceleration. To obtain higher accelerations speeds on land: 1228 km h–1. The rockets than about 10 m s–2, one has to circumvent the also allow for very high acceleration: 0 to road grip issue, e.g. by using a jet engine instead of 965 km h–1 in 16 s. traction through the tires (see Fig. 5.13 and exercise 7). 108 CHAPTER 5 Example 5.11 Decelerating car, graphically Velocity, 𝑣 Fig. 5.14 shows the (𝑡, 𝑣) −graph of a car approaching a red light, i.e. the speedometer reading as a function of time. As you can see, the car at first drives at constant velocity (until 𝑡1 ), and then the speed decreases until 𝑡1 𝑡2 𝑡3 𝑡4 Time, 𝑡 the car has stopped in front of the red light Figure 5.14 The car slows down to stop at a red light. at time 𝑡4. More specifically:  Before 𝑡1 , and again after 𝑡4 , the velocity graph is horizontal. This means constant velocity and hence zero acceleration.  Between 𝑡1 and 𝑡4 the velocity is changing, so in this time interval the car accelerates (negative acceleration because the gradient of the curve is negative).  From 𝑡2 to 𝑡3 the velocity graph is a straight line, so in this time interval the acceleration is constant. Example 5.12 Acceleration of gravity: 𝑔 = 9.8 m s −2 The acceleration of gravity is the acceleration experienced by a freely falling object when air resistance is negligible. Its value is easy to determine experimentally. If you do not go too many kilometers away from the surface of the Earth, you will find 𝑔 = 9.8 m s–2. This means that the speed of a falling object increases by 9.8 m s–1 every second. After 2 s, the speed will have increased by 19.6 m s–1 since 19.6 is 2 times 9.8. More generally, after the time 𝑡 the speed will have increased by 𝑔 × 𝑡. Most accelerations in our everyday life are considerably smaller than the acceleration of gravity. The acceleration of the cars in Example 5.9 were about 20 % and 10 % of 𝑔, respectively. In an amusement park ride, you are likely to experience accelerations larger than 𝑔. And in a car crash, the speed of the car drops from a large value to zero in a very short time span, and therefore the acceleration is very large – and possibly fatal. In the subatomic world of electrons and other charged particles, however, accelerations many thousand times larger than our everyday accelerations are commonplace. Example 5.13 Crashing car km If a car collides head on with a tree or another obstacle, the speed of the car drops from, say, 70 to h zero in a very short time, typically ∆𝑡 = 0.2 s. This gives an average acceleration of 70 ∆𝑣 3.6 m s −1 𝑎= = = 97.2 m s −2 = 100 m s−2 (1 sig. fig. ) ∆𝑡 0.2 s This is roughly 10 times the acceleration of gravity. Your internal organs will rip from where they ought to be and your brain will collide with the inside of your skull like a tomato thrown at a wall. Seat belts and air bags will increase ∆𝑡 a little bit, making the acceleration somewhat less – and you may survive.  MOTION 109 On YouTube you can find high speed camera recordings of car manufacturers’ controlled car crashes. Often time and position can be read off the video, and then you can calculate the acceleration. Try it! 5.5 Motion at constant acceleration (optional section) In section 5.3 we looked at motion without acceleration. The velocity was constant, and we derived the formula 𝑥 = 𝑣 𝑡 + 𝑥0 for the position as a function of time. We saw how the (𝑡, 𝑥)-graph is a straight line. Constant acceleration… When there is constant acceleration, the velocity is no longer Then the position graph is a parabola! constant, and it turns out that the position depends quadratically on time. The (𝑡, 𝑥)-graph is a parabola: 1 𝑥 = 2 𝑎𝑡 2 + 𝑣0 𝑡 + 𝑥0 Here 𝑥 stands for position, 𝑎 is the acceleration (which must be constant), 𝑡 is the time measured since the position was zero, 𝑣0 is the initial velocity with which the object started at time zero, and 𝑥0 the initial position (plug in 𝑡 = 0 into the equation). This relatively complex formula is derived in exercises 25+27. 1 Notice the important special case where initial velocity and position are both zero: 𝑠 = 2 𝑎𝑡 2. Example 5.14 Constant acceleration Position / m Velocity / m s–1 If a car starts at speed zero and accelerates by three meters per second every second, then its velocity and position can be calculated by 𝑣 = 𝑎 𝑡 = (3.0 m s −2 ) × 𝑡 1 𝑥 = 2 𝑎 𝑡 2 = (1.5 m s −2 ) × 𝑡 2 respectively. During the first three seconds, the velocity increases to Time / s Time / s (3.0 m s −2 ) × (3.0 s) = 9.0 m s−1 Figure 5.15 Velocity and position when accelerating and the position to from rest at constant acceleration 3 m s–2. (1.5 m s −2 ) × (3.0 s)2 = 13.5 m To find out how long it takes the car to drive, say, 6.0 m, one can either look at the position graph (red 1 dotted lines in Fig. 5.15 show that it takes 2.0 s) or solve the position equation 𝑥 = 2 𝑎𝑡 2 for 𝑡: 2𝑥 2×6.0 m 𝑡 = √ 𝑎 = √3.0 m s−2 = 2.0 s 110 CHAPTER 5 Example 5.15 The area under a velocity graph It is generally true that the displacement of an object velocity during a certain time interval can be calculated as the area under its (𝑡, 𝑣) − graph for that time interval. 𝑎𝑡 1 Justification for this important fact can be found in Area: 𝑎𝑡 2 2 exercise 25. Let us therefore look at the area under the (𝑡, 𝑣)-graph in the special case where an object accel