PHYS 1211 - Energy and Environmental Physics

1. PHYS 1211 - Energy and Environmental Physics Michael Ashley Lecture 2 Mechanical Energy

2. This Lecture • A Bit of History • Energy and Work • Units of Energy • Power • Kinetic and Potential Energy

3. 2000 1700 1800 1900 1600 Isaac Newton developed his system of mechanics entirely without the concept of Energy. Newton’s description was in terms of forces and momentum and the effects forces had on the motion of objects.

4. 2000 1700 1800 1900 1600 Newton’s contemporary (and rival) Gottfried Leibniz introduced the idea of what he called “vis viva” (living force). “vis viva” = mv2and was conserved in some interactions between particles. It is what we now know as kinetic energy (although we now use 1/2 mv2) The term “vis viva” continued to be used for mechanical energy up to the 1850s. But Leibniz’s work was largely ignored because it was thought to be incompatible with Newton’s law of conservation of momentum.

5. 2000 1700 1800 1900 1600 It was clear that “vis viva” was not always conserved. At the end of the 18th century several scientists (such as Antoine Lavoisier and Pierre-Simon Laplace) begin to suspect that the lost energy appears as heat. In 1798 Count Rumford (Benjamin Thompson) studied the frictional heat produced in boring the barrel of a cannon. The mass of the cannon does not change as it is heated. An indefinite amount of heat can be generated by friction. Heat could not be a substance — as argued by the then widely accepted “caloric” theory of heat. Heat must be “a form of motion”

6. 2000 1700 1800 1900 1600 In 1807 Thomas Young first used the term “energy” in its modern scientific context. In the 1840’s James Prescott Joule carries out a series of experiments, showing the equivalence of mechanical energy and heat. When a certain amount of work is done (e.g. by a falling weight), a corresponding amount of heat is produced. But Joule’s ideas still met resistance from supporters of the “caloric” theory.

7. 2000 1700 1800 1900 1600 In the 1850s scientists including Heinrich Helmhotz, William Thompson (Lord Kelvin) and Rudolf Clausius formulate the laws of thermodynamics. They show that energy exists in forms such as mechanical energy, heat, light, electricity and can be converted between these forms but can never be created or destroyed (the law of conservation of energy). In just a couple of decades from about 1840-1860 “energy” develops from a largely unknown term to one of the fundamental concepts of Physics. William Thompson (Lord Kelvin)

8. 2000 1700 1800 1900 1600 By 1930 scientists are so convinced of the law of conservation of energy, that when it appeared not to be conserved — in the process of beta decay — Wolfgang Pauli proposes a new particle “the neutrino” to explain where the energy has gone. 26 years later the existence of the neutrino was experimentally confirmed — an experiment that won Frederick Reines the Nobel Prize in physics. Bubble chamber tracks showing the first detection of a neutrino. The invisible neutrino hits a proton and produces three tracks (a mu meson, a pi meson and the proton).

9. What is Energy? • Energy is defined as the “capacity to do work”. • So what is work? • In Physics work is defined as the product of a force times the distance through which the force acts. • W = F d

10. The Joule • The S.I. unit of work (and energy) is the Joule (J). • Named after James Prescott Joule (1818-1889), the British physicist who studied the relationship between work and heat. • One Joule is the work done when a force of one Newton moves through a distance of one metre.

11. Energy and Work • Energy and Work are both measured in Joules so what is the difference between them? • Work is the action involved when a force acts on a system. • Energy is a property of the system. • We “do work on” a system and the energy of the system increases.

12. Energy and Work • Work is only being done for the time for which the force acts. • But the Energy gained by the system continues to exist — at least until something else happens to change the state of the system again.

13. People Doing Work Rock climbing — lifting the climbers weight. Lifting weights — work is done on the weight. Ball games such as cricket work is done on the ball.

14. Machines doing work A crane lifting a load A rocket launch A car accelerating

15. Example Hossein Rezazdeh (“the world’s strongest man”) holds the world record in the super-heavyweight weightlifting class and won gold medals at the Sydney and Athens Olympics. His record for the “clean and jerk” is 263.5 kg. How much work is he doing in such a lift? The gravitational force is mg = 263.5 * 9.81 = 2585 Newtons. Assume the distance the weight is lifted is 2m. The work done is force x distance = 2 x 2585 = 5170 Joules. = 5.17 kJ

16. How much is a Joule? “AA” alkaline battery 10,800 J 10.8 kJ Tim Tam biscuit 400,000 J 400 kJ A litre of unleaded petrol 34,800,000 J 34.8 MJ Average daily household electricity usage 72,000,000 J 72 MJ

17. Other Work (Force x distance) Energy Units • Erg • The unit on the old cgs system. An erg is the work done when a force of one dyne moves through one cm. 1 erg = 10–7 J • Foot pound (ft-lb) • Imperial unit of work. A force of one pound force moving through one foot. 1 ft-lb = 1.356 J

18. “Heat” Energy Units • 1 calorie is the energy needed to heat 1 gram of water through 1 degree C. 1 cal = 4.184 J • 1 kcal (or food Calorie) = 1000 calories = 4.184 kJ. • 1 Btu (British Thermal Unit) is the energy needed to heat 1 lb of water through 1 degree F. 1 Btu = 1055 J • Despite the name now mostly used in the USA.

19. Other Units • 1 kilowatt-hour (kWh) is the energy corresponding to 1 kW of power used for 1 hour (since 1 kW = 1000 J/sec and 1 hour = 3600 sec, 1 kWh = 3.6 x 106 J). • Tonne of oil equivalent (toe) is often used in statistics of national and global energy usage. • 1 toe = 41868 MJ = 4.1868 x 1010 J (according to IEA convention).

20. Power • Power is defined as the rate of doing work or converting energy. • The SI unit of power is the Watt (W) • Named after Scottish inventor James Watt who made major improvements to the steam engine. • One Watt is one Joule per second.

21. Desktop Computer (iMac 20”) 200 W Electric Heater 2000 W Small Car (Honda Jazz) 61 kW Formula 1 Car 550 kW Queen Mary 2 86,000 kW

22. Other Power Units • A horsepower (hp or HP) is the unit James Watt actually used to measure power. • When steam engines were first introduced this was a useful unit as it indicated how many horses the engine could replace. • One hp = 746 W. • Horsepower is still sometimes used to describe the power of engines.

23. Energy Units & Conversion Factors Prefixes: Micro  10-6 Giga G 109 Milli m 10-3 Tera T 1012 Kilo k 103 Peta P 1015 Mega M 106 Exa E 1018 Energy Units 1 Btu = 1055 J = 252 cal [British Thermal Unit] 1 cal = 4.184 J [calorie] 1kcal = 1000 cal = 1 food Calorie 1kWh = 3.6 x 106 J = 3413 Btu [Kilowatt hour] 1 Quad = 1015 Btu = 1.055 x 1018 J 1 GJ = 109 J = 948,000 Btu Power Units 1 W = 1 J/s = 3.41 Btu/hr [Watt] 1 hp = 2545 Btu/hr = 746 W [Horse Power] Fuel 1 barrel crude oil = 5.8 x 106 Btu = 6.12 x 109 J 1 standard ft3 natural gas (SCF) = 1000 Btu = 1.055 x 106 J 1 therm = 100,000 Btu 1 ton bituminous coal = 25 x 106 Btu 1 ton 238U = 70 x 1012 Btu 1 ton = 907.2 kg 1 metric ton = 1 tonne = 1000kg

24. Electrical Energy and Power • The power of an electrical appliance is the product of the voltage (V in volts) and the current (I in Amps) flowing through it: Power = VI I = 2 Amps Mains Supply 240 V Power = VI = 240  2 = 480 Watts

25. Energy in a Battery • Energy content of a battery is usually quoted in ampere-hours (Ah). • Consider a 6 V battery with a capacity of 10 Ah • It can deliver 60 W for 1 hour (=3600 seconds) • Energy = 60  3600 J = 216,000 J = 216 kJ Electrical Energy will be discussed in more detail later in the course.

26. Kinetic Energy • The Kinetic Energy of an object is the energy it possesses because of its motion at a velocity v. KE = 1/2 mv2 • This expression can be derived from Newton’s 2nd Law. F = ma

27. Kinetic Energy • Consider an object accelerating from rest with constant acceleration a because of a force applied to it. After time t: v = at s = 1/2 at2 So t = v/a s = 1/2 a(v/a)2 = 1/2 v2 /a Work = Fs = 1/2 F v2/aand F = ma = 1/2 mv2 • So the work done by a force accelerating an object is 1/2 mv2and this must be the energy gained by the accelerating object. s F m m t = 0 v =0 t = t v = at

28. Potential Energy • Potential Energy is the energy of an object by virtue of its position. • Gravitational potential energy is energy due to its position in the gravitational field. Since the force of gravity on an object is: F = mg where g is thegravitational acceleration(9.81 ms–2) then: PE = Force x Distance = mgh where h is the height of the object.

29. Work, PE and KE • When work is done on an object its potential or kinetic energy (or both) is changed. • For example: • Lifting a weight — the potential energy of the weight is increased. • Throwing a ball — the kinetic energy of the ball is increased.

30. Law of Conservation of Energy • If we consider just mechanical energy then the following relations hold: E = PE + KE Total Energy = kinetic + potential energy W = E = KE + PE Work done on a system changes its total energy If no external work is done on a system: E = 0 KE = –PE i.e. Total energy cannot change, but energy can change from kinetic to potential or vice versa.

31. Elastic Potential Energy • Another type of potential energy is the energy stored in a spring or elastic material. • The force due to a spring is kx. k is the spring constant and x is the amount the spring is compressed (Hooke’s Law) • The PE is then given by: PE = 1/2 kx2

32. Potential Energy as a Power Source We can use potential energy to power machines. For example Clocks: Weight driven clocks use the gravitational potential energy of a falling weight to drive the clock mechanism. Wind-up spring-driven clocks use elastic potential energy stored in a spring.

33. Exchange of Potential and Kinetic Energy • Many systems involve exchange of potential and kinetic energy. • Simplest example is a falling object (in the absence of air resistance). • After falling a vertical distance s • PE = –mgs (–ve sign since PE is lost) • And it gains an equal amount of KE • 1/2mv2 = mgs • v = (2gs) • We could have got the same result using Newton’s 2nd Law — using energy is an alternative approach to such problems.

34. Roller Coaster A roller coaster operates by converting energy between potential energy and kinetic energy.

35. Roller Coaster Initial Position (max PE) As energy is lost due to friction, successive peaks have to be lower or the car would not have enough energy to reach them Gaining velocity and KE Minimum PE, maximum KE and velocity

36. Escape Velocity • Escape velocity is the velocity needed for a spacecraft to completely escape the Earth’s gravitational field. • Spacecraft sent to other planets (e.g. Mars) need to reach escape velocity. NASA’s Mars Reconnaissance Orbiter

37. Escape Velocity • Previously we have used the expression mgh for PE, but this is only correct near the surface of the Earth where the gravitational acceleration has the fixed value g (= 9.81 ms–1). • In general we have to use the full form of Newton’s law of gravity. F = GMm/r2 Where M is the mass of the Earth, r is the distance from the Earth’s centre, and G is the gravitational constant (G = 6.67  10–11) • From F = ma we can now see that the acceleration due to gravity is GM/r2

38. Escape Velocity PE = 0 at r =  • So in the expression for PE (mgh) we need to replace g with GM/r2 and h with r giving: PE =–GMm/r The – sign is needed so that energy increases upwards. • At infinite distance (r = ) from the Earth PE = 0. • At the surface of the Earth (r = R) PE = –GMm/R • To launch a spacecraft so it escapes from the Earth we need to give it a KE (= 1/2mv2) equal to the PE change from the surface to infinite distance. 1/2mv2 = GMm/R v2 = 2GM/R PE = –GMm/R Earth Radius R Mass M

39. Escape Velocity • For Earth: • M = 5.97  1024 kg • R = 6.37  106 m • So: v2 = v = 11,181 ms–1 = 11.2 km s–1 = 40,300 kph. 2  6.67  10–11 5.97  1024 6.37 106 Soyuz-Fregat rocket launching the ESA Venus Express spacecraft

40. The Pole Vault • The pole vault illustrates an efficient way of converting kinetic energy into potential energy.

41. The Pole Vault The run up — the pole vaulter must run up as fast as possible. World record holder Sergey Bubka has been measured at 22.2 mph = 9.93 ms–1. His kinetic energy is then: KE = 1/2mv2 = 1/2 80  9.932 KE = 3944 J (for mass m = 80 kg)

42. The Pole Vault As the vaulter plants the pole, he starts to rise (converting some of his KE to PE), but also bends the pole. Much of the original kinetic energy is now stored as elastic potential energy in the pole.

43. The Pole Vault As the pole starts to straighten it releases its stored elastic potential energy and converts this to gravitational potential energy of the vaulter.

44. The Pole Vault The pole is now straight and the vaulter has reached his maximum height. All the original KE should now be converted to the vaulter’s PE. PE = mgh = original KE = 3944 J h = 3944/(mg) = 3944/(80  9.81) h = 5.03 metres.

45. The Pole Vault • But Sergey Bubka’s world record was 6.14m and we have calculated only 5.03m - How is this possible? • In fact he starts off with his centre of mass already at a height of about 1 metre. • When he crosses the bar his centre of mass would be only slightly above the bar. • In addition he can make an extra push off the pole at the top of the swing. • However, it would seem like his performance is near the limit of what is physically possible. • Perhaps not surprising that his 1994 world record was not broken until 2014 (6.16m Renauld Lavillenie).

46. The 100m Sprint The 100m event is often referred to as the Blue Riband event of the Olympics and the men who run it as the “fastest men on Earth”.

47. The 100m Sprint An olympic sprinter does the 100m run in about 10s (world record is 9.58s). This means an average speed of about 10 ms–1 However the top speed is more like 12 ms–1 (26.95 mph). The plot at right shows that in the first 20m the sprinter has accelerated to about 10 ms–1 (22mph). Using: v2 = 2as and v = at We can calculate that: a = 2.5 ms–2 t = 4 s i.e. the sprinter takes about 4 seconds to run the first 20m accelerating at 2.5 ms–2 to 10 ms–1 Then KE = 1/2 mv2 = 4000 J Power = 4000/4 = 1000 W ( = 1.34 horsepower)

48. Tour de France Climb of the Cime de la Bonnete: Starts at 1152m Ends at 2802m A climb of 1650m The race leaders completed this climb in 69 minutes. Speed = 6.4 ms–1 (compared to ~12 ms–1 on flat stages) For m = 80kg (cyclist+bike): PE =mgh = 80  9.81  1650 = 1,295,000 J = 1.29 MJ Power = 1295000/(69 60) = 313 W This probably underestimates total power as it only accounts for PE gain - not air drag etc. Stage 16 of the 2008 Tour de France

49. Power Output of Human Body • So while about 1000W can be achieved over short periods, for long duration events power outputs are more like 400 W maximum. • We will look at the reasons for this difference later — but briefly. • In endurance events power is limited by the ability of the cardiovascular system to supply O2 to the muscles. • Over short periods a different process (“anaerobic respiration”) can be used to supply energy at higher rates.

50. Next Lecture • We will continue our introduction to energy by looking at thermal energy.