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AS Physics Unit 10 Work, Energy, Power Mr D Powell Chapter Map Spec Link: 3.2.1 Principal of cons of energy, applied to examples involving GPE, KE, and Work Done against resistive forces Mr Powell 2009 Index 10.1 Work and energy Specification link-up 3.2.1: Work, energy and power What is energy and how do we measure it? Does energy ever disappear? What is meant by work (in the scientific sense)? Mr Powell 2009 Index What is work? Work is the energy transfer that takes place when a force causes an object to move. work done = force applied × distance moved in direction of force What is crucially important is to realise that no energy transfer takes place when the motion is not in the direction of the force! work done is measured in joules (J) force is measured in Newtons (N) distance is measured in metres (m) 100 N W = Fs = 100N x 5m = 500Nm = 500J 5m Mr Powell 2009 Index Work done by a force at an angle The same box is now dragged by a rope, which is raised at an angle θ to the horizontal. F θ This time, the box moves in a different direction to the direction of the applied force. How does this affect the work done? Can you think of any suggestions? Mr Powell 2009 Index Calculating work done at an angle F When calculating the work done by a force acting at an angle, it is useful to break the force down into components. θ The tension in the rope can be broken down into a horizontal and a vertical component. The vertical component does no work because the box does not move in that direction. F Fsinθ θ So to calculate work done by a force at an angle: Fcosθ work done = force in direction of movement × distance moved W = Fscosθ Mr Powell 2009 Index Practical F Using a force sensor can you confirm this idea by conducting a simple practical? θ Take care to control your experiment and all variables apart from the angle. You may require… • • • • Force sensor String Block Weights F Fsinθ θ Fcosθ work done = force in direction of movement × distance moved W = Fscosθ Mr Powell 2009 Index Quick Question.... 10 N A toy car is pulled along by a piece of string which is at 30° to the horizontal. 1) Calculate the work done in pulling the toy if the tension in the string is 10 N, and it is pulled along 5 m. 2) Comment on why the actual energy input will need to be more than this? 3) Comment on why power delivery to the car to make it move at a steady speed will not be uniform? 30° 5m W = Fscosθ = 10 × 5 × cos30° = 43.3 J 2) We will need to overcome frictional forces as we move so energy will be wasted 3) The friction forces will be higher initially until the car is moving when this static friction is overcome the dynamic friction is lower. Hence, power will need to change. Mr Powell 2009 Index Force Distance Graphs Questions… Using… work done = force applied × distance moved in direction of force & WD = Fscos We can now relate the energy transfer for a box to an area under the graph. Mr Powell 2009 Index Force Distance Graphs Questions… Using… work done = force applied × distance moved in direction of force What is the energy transfer in Joules in Each of these cases if the force was applied at 45 in the left picture and 60 in the right case? 27J x cos 60 =13.5J 100J x cos 45 = 71J Mr Powell 2009 Index 10.2 Kinetic energy and potential energy Specification link-up 3.2.1: Work, energy and power When an object is lifted, what happens to the work done on it? If it is then allowed to fall, what energy change takes place? What is the effect on the kinetic energy of a car if its speed is doubled? Mr Powell 2009 Index What is kinetic energy? Kinetic energy (KE or Ek) is the energy of an object due to its speed.... kinetic energy = ½ × mass × speed2 Ek = ½mv2 You should try and get in the habit of using the subscript “k” to show kinetic every time you quote energy from now on. Also EP for dealing with potential energy or it will get very confusing. kinetic energy is measured in joules (J) mass is measured in kilograms (kg) speed is measured in metres per second (ms-1). Eureka Video Mr Powell 2009 Index Deriving Ek = ½mv2 Consider a force F acting on an object of mass m, initially at rest, moving it a distance s in time t. Using the SUVAT equations & Newton's 2nd law can you derive kinetic energy? From ‘suvat’ equations: s = ½ (u + v)t a = (v – u) / t Because u = 0 ms-1: s = ½vt a=v/t Newton’s 2nd law: F = ma Substituting a = v / t: F = mv / t Work done by force: W = Fs W = (mv / t) × ½vt W = ½mv2 Work done = energy transferred: Ek = ½mv2 Mr Powell 2009 Index What is gravitational potential energy? Gravitational potential energy (GPE, Ep or Egrav) is the energy of an object due to its position in a gravitational field. The Ep gained by a mass is proportional to the force used to lift it, and the distance it is lifted. It is often talked about in terms of a change in an object’s Ep due to a change in its height: ΔEp = mgΔh ΔEp = mgΔh = 0.250kg × 9.81Nkg-1 × 2m Example... A supermarket employee lifts a baked bean tin, weighing 250 g, from the floor, to a shelf 2 m high. How much gravitational potential energy does it gain? (g = 9.81 N kg-1) = 4.9Nm = 4.9 J Mr Powell 2009 Index Ek and Ep If resistive forces, such as friction and air resistance, are ignored, Ek and Ep are related as follows: loss of Ek = gain in Ep or lose of Ep = gain in Ek For example, if an object of mass m is released above the ground at height h, it will gain speed, v, as it falls. Due to the conservation of energy, and assuming air resistance is negligible, after falling a height of Δh: ½mv2 = mgΔh Mr Powell 2009 Index Exam Question - Conservation of energy A ball of mass 400 g is thrown upwards at a speed of 5 ms-1. (g = 9.81 N kg-1). a) What is the ball’s Ek as it is released? Ek = ½mv2 = ½ × 0.4 × 52 =5J b) What is the ball’s maximum gain of Ep? ΔEp = Ek =5J c) What is the ball’s maximum height? Ep = mgh h = Ep / mg = 5 / (0.4 × 9.81) = 1.27 m Mr Powell 2009 Index Pendulum – Gains & Loses KE / GPE Pendulums gain and lose PE & KE as the move. As you change height from equilibrium position (blue line) you will gain or lose PE energy according to the formulae.. ΔEp = mgΔh If we lose energy on the journey ΔEp = mg(ho –h) Where h0 is the original height and h is the height h is the height the pendulum swings to on the other side. (lower than original) Mr Powell 2009 Index 10.3 Power Specification link-up 3.2.1: Work, energy and power Which physical quantities are involved in power? (AO1a) How could you develop more power as you go up a flight of stairs? (AO2b) Why is a 100 W light bulb more powerful than a 40 W light bulb when each works at the same mains voltage? (AO2a) Mr Powell 2009 Index Making things Easy? Mr Powell 2009 Index What is power? Power is the rate at which work is done, or the rate at which energy is transferred. power = work done / time taken P = W/t power is measured in watts (W) work done or energy transferred is measured in joules (J) time is measured in seconds (s). Question... A crane lifts a load of 1500 kg a height of 25 m at a steady rate, in a time of 2 min. What is the power of the crane? W = energy transferred = ΔEp ΔEp = mgΔh = 1500 × 9.81 × 25 P=W/t = 367 875 / 120 = 3070 W = 367 875 J Mr Powell 2009 Index Key Points…. If we think of a machine at work that lifts an object….. Work Done or W = Fs However, if the object is moving at a constant velocity we can make a simple substitution to eliminate “s” from the equation… s = 0.5(v+u)t s = vt (const velocity) Pout Fv Power W Js-1 Force N Velocity ms-1 useful Eff total Hence…. Work Done is Fs = Fvt Power is then worked out as Fvt / t = Fv Mr Powell 2009 Index Motive power Example The power outputted by a powered object, such as an engine or muscles, is sometimes called the motive power. If the powered object is moving at a constant speed at a constant height: power = force × speed P = Fv At constant speed and height, the force produced by the powered object is equal but opposite to all resistive forces acting on the object, such as friction and air resistance. Question... What is the resistive force on a cyclist who has leg muscles of power 200 W each and who reaches a top speed of 10 ms-1 on a level road? P = Fv F =P/v = (200 × 2) / 10 = 40 N Mr Powell 2009 Index Power: efficiency Efficiency is the ratio of useful work done by a device, to the total work done (or the ratio of useful output energy to the total energy input). Question... For example, what is the efficiency of a 60 W filament lamp that gives out 1 W of light? efficiency = 1W / 60W efficiency = useful work done / total work done efficiency = useful energy output / total energy input = 0.017 = 1.7% Efficiency is often expressed as a percentage. Efficiency is always less than 100%, as no device is perfect and some energy is always lost. Mr Powell 2009 Index 10.4 Energy & Efficiency 1. How can power be linked to velocity 2. Into what form of energy is wasted energy almost always converted? 3. Can any device ever achieve 100% efficiency? How Escalators Work Outcomes ….. Be able to complete calculations to work out the output Power of… (AO2b) 1. 2. 3. 4. an electric motor (A-E) A pulley system (A-E) A winch (A-D) (involves efficiency calcs) An Escalator (Higher Level) (A-C) Mr Powell 2009 Index Key Points…. If we think of a machine at work that lifts an object….. Work Done or W = Fs However, if the object is moving at a constant velocity we can make a simple substitution to eliminate “s” from the equation… s = 0.5(v+u)/t s = vt (const velocity) Pout Fv Power W Force N Velocity ms-1 useful Eff total Hence…. Work Done is Fs = Fvt Power is then worked out as Fvt / t = Fv Mr Powell 2009 Index Independent Study… Work through the first two examples on page 155 in the text book on the Electric Motor & Pulley System For each one… 1. 2. 3. 4. Draw a diagram Write out the quantities Write out the formulae you might require Work through the calculations step-by-step Higher Level Now turn to page 156 and work through the calculations on the winch in the same way. Move your work onto a higher level with an efficiency calculation Final Stage If you are not so confident work through Summary Q1/2 then 3/4. If you are confident do Q3/4 and then ask for a worksheet. Mr Powell 2009 Index Summary Q Extra Help… Q1) a) Useful energy per second = Power Power = Fv = 30N x 15ms-1 =450Js-1 b) Useful energy is 450Js-1 which is only 25% of the input energy. Hence = 450Js-1 x4 = 100% of the energy =1800 Js-1 Q2) a) E = Pt E = 60Js-1 x 8s E =480J b) E = mgh E = 20N x 2.5m E = 50J c) Efficiency = = 50J / 480J = 0.1 = 10% Mr Powell 2009 Index Summary Q Extra Help… 3) a) Efficiency of 35% gives us 200MW. Means that you need to produce internally before the waste Input Power = 1/0.35 * 200MW = 571.4MW = 570MW (2 s.f.) If each kg of fuel produces 80MJkg-1 Your would need 571.4MW / 80MJkg per second = 7.1425kgs-1 24 hours is 60 x 60 x 24 = seconds in a day. Total burn in a day = = 7.1425kgs-1 x 60 x 60 x 24 =617112kg = 6.2 x 105kg Mr Powell 2009 Index Summary Q Extra Help… 4) Pout = 6.2kW a) Fuel Burn 45MJkg-1 Vehicle travels 18km on one kg of fuel What is the time taken to travel 18km at 30ms-1 when burning fuel at rate of 1kg per 18kg or 45MJ of energy per 1kg t = s/v = 18,000m / 30ms-1 = 600s b) E = Pt = 6.2kW x 600s = 3720000J = 3.72MJ c) Overall Efficiency = Input Power / Useful Power = 3.72MJ / 45MJ = 0.083 = 8% Mr Powell 2009 Index Does an escalator run at constant power? Facts about a typical escalator How Escalators Work slopes at 30° to the horizontal each step rises 15 m in 1.0 minute efficiency of electric motor is 70% When fully laden 80 people get on and off in 1.0 min average mass of one person 70 kg a frictional force of 1.2 × 104 N, moving 30 m in 1.0 min, acts against the motion 1) Calculate the potential energy given each minute to the people on a fully laden escalator. Take g = 9.8 N kg−1. PE = mgh = = 80 × 70kg × 9.8 N kg−1 × 15m = 823 200 = 820 000 J (2 s.f.) Mr Powell 2009 Index Does an escalator run at constant power? Facts about a typical escalator How Escalators Work slopes at 30° to the horizontal each step rises 15 m in 1.0 minute efficiency of electric motor is 70% When fully laden 80 people get on and off in 1.0 min average mass of one person 70 kg a frictional force of 1.2 × 104 N, moving 30 m in 1.0 min, acts against the motion 2) Show that the kinetic energy given to the people on the escalator is small compared to the potential energy they receive. (Hint: find the speed up the slope.) 15 m rise up a 30° slope is 30 m along the slope… Distance travelled is … hyp = 15m/sin 30 = 30m speed up slope = 30m/60s = 0.5 m s−1 KE of all people on escalator = 0.5mv2 = 80 × 0.5 x 70kg × (0.5ms-1)2 = 700 J Mr Powell 2009 Index Does an escalator run at constant power? Facts about a typical escalator How Escalators Work slopes at 30° to the horizontal each step rises 15 m in 1.0 minute efficiency of electric motor is 70% When fully laden 80 people get on and off in 1.0 min average mass of one person 70 kg a frictional force of 1.2 × 104 N, moving 30 m in 1.0 min, acts against the motion 3) Calculate the work done per second by the frictional force. P = Fv = 1.2 × 104N ×30m/60ms-1 = 6000Js-1 = 6000 W Mr Powell 2009 Index Does an escalator run at constant power? How Escalators Work Facts about a typical escalator slopes at 30° to the horizontal each step rises 15 m in 1.0 minute efficiency of electric motor is 70% When fully laden 80 people get on and off in 1.0 min average mass of one person 70 kg a frictional force of 1.2 × 104 N, moving 30 m in 1.0 min, acts against the motion 4) Using your answers to questions 1 and 3, calculate: a) the total output power of the electric motor b) the input power to the electric motor c) the overall efficiency of the escalator. a) Total output power = increase in PE/sec + WD/sec against friction = 823200J/60s + 6000Js-1 = 13 720Js-1 + 6000Js-1 = 19 720Js-1 = 20 000 W (2 s.f.) Mr Powell 2009 Index Does an escalator run at constant power? How Escalators Work Facts about a typical escalator slopes at 30° to the horizontal each step rises 15 m in 1.0 minute efficiency of electric motor is 70% When fully laden 80 people get on and off in 1.0 min average mass of one person 70 kg a frictional force of 1.2 × 104 N, moving 30 m in 1.0 min, acts against the motion 4) Using your answers to questions 1 and 3, calculate: a) the total output power of the electric motor b) the input power to the electric motor c) the overall efficiency of the escalator. b) Efficiency of electric motor is 70% or = 0.7 = useful power output/total input power = 19720/ input power Rearrange… input power = 19720/0.7 = 28 171 ≈ 28.2 kW (3 s.f.) Mr Powell 2009 Index Does an escalator run at constant power? How Escalators Work Facts about a typical escalator slopes at 30° to the horizontal each step rises 15 m in 1.0 minute efficiency of electric motor is 70% When fully laden 80 people get on and off in 1.0 min average mass of one person 70 kg a frictional force of 1.2 × 104 N, moving 30 m in 1.0 min, acts against the motion 4) Using your answers to questions 1 and 3, calculate: a) the total output power of the electric motor b) the input power to the electric motor c) the overall efficiency of the escalator. c) Overall efficiency of escalator = useful power output/ total input power = 13720Js-1/28171Js-1 (increase PE/sec Q4a / From Q4b ≈ 0.50 or 50% efficient Mr Powell 2009 Index 10.5 Renewable Energy Specification link-up 3.2.1: Conservation of energy 1. How much power can we generate from different renewable energy resources? 2. Could we meet all our energy needs from renewable energy resources? 3. Can we meet our demand for energy and cut carbon emissions at the same time? AO1a/AO2b – Work out energy transfers for a Solar Panel (Basic) 1 2 v V 2 AO1a/AO2b – Work out the energy transfers for a Tidal Power (Medium) AO1a/AO2b – Work out the Power of a Wind Turbine (Harder) Mr Powell 2009 Index Crystalline Cells There are two types of solar cell. These are mono crystalline and poly crystalline. These are wired in series to produce solar panels As each cell produces a voltage of between 0.5 and 0.6 Volts, 36 cells are needed to produce an open-circuit voltage of about 20 V. This is sufficient to charge a 12 Volt battery under most conditions. A solar panel of area 1m2 in space would absorb energy from the sun at a rate of 1400Js-1 if all the incident energy went into the cell However, at the surface of the Earth some of this energy is absorbed by the atmosphere. Water heating panels may absorb 500Js-1 and heat water to around 70C in a British Summer The output from a solar cell is at its greatest when the light hits the cell at right angles. They are made of a thin sandwich of Cover glass, Transparent glue, Anti reflective coating then Silicon and gallium arsenide conductor layers to generate current flow Mr Powell 2009 Index Solar Cell efficiencies http://en.wikipedia.org/wiki/Solar_cell See how they efficiencies change by structure and year of development! Mr Powell 2009 Index How does it work…. Imagine a Power Station has area of 10km x 10km (1 x108m2) and traps a tide of 6m depth each day… You would exchange a volume of water of.. 1 x 108m2 x 6m = 6 x 108m3 With a density of 1000kgm-3 If the water drops a height on release of 1m then energy released to turn a turbine would be… E = mgh = 6 x 108m3 x 1000kgm-3 x 1m x 9.81Nkg-1 = 5.886 x 1012J E = 5.9 x 1012J 1 m2 = 0.000001 (or 1 x 10−6) km2 Mr Powell 2009 Index Other Methods... Mr Powell 2009 Index Wind Issues/Facts • The larger the blades or wind the more energy produced. • Energy only when there is wind. • Low running costs • Cause of noise pollution • Cause of visual pollution KE of wind KE per unit volume KE per unit of a cylinder of air swept out by turbine Volume of cylinder is speed x area of cone KE per second or Power as cylinder of unit volume area A passes through turbine.... 1 2 KE mv 2 1 2 KE / V v 2 1 2 KE v V 2 V vA 1 2 Power v vA 2 1 3 Power v A 2 NB: They will give you help on this in the exam! Mr Powell 2009 Index Summary Question Extra Help… 1) 2) Max Power v3 Solar panel of areas 1m2 produces 200W. 1.2Mw when v = 10ms-1 Hence we can say that is 200Wm-2. When speed is 15ms-1 To produce this much we need …. v3 = (10ms-1)3 = 1000 v3 = (15ms-1)3 = 3375 200,000,0000Mw / 200Wm-2 = 1 x 107m2 This is 3.375x more. Hence Power … = 3.375 x 1.2 = 4.05MW = 4MW Mr Powell 2009 Index Summary Question Extra Help… 3) Power Station Efficiency = 25% = 0.25 h = 650m Volume water passing /s produces 200MW of power. Density of water = 1000kgm-3 What volume of water would require this… 200MW is 25% of Input energy. Hence energy generated by the fall is 800MW. Hence drop E = mgh E / hg = m m = 800,000,000W / (650m * 9.81) = 125.5 x 103kg m/d = volume = 125.5 x 103kg / 1000kgm-3 v = 125m3 Mr Powell 2009 Index Summary Question Extra Help… 4) Water trapped by tidal station is 200km2. Tide or h = 3m. Water released 6 hours. a) Mass of water is…. volume x density m = area x depth x density m = 200 x 106 m2 x 3m x 1050kgm-3 m = 6.3 x 1011 kg b) This mass is then released over 6 hours t = 60 x 60 x 6s = 21,600s E = mgh / t E = 6.3 x 1011 kg x 9.81 Nkg-1 x 3m / 21600s E = 860MW But Average PE is ½ this… E = 430MW Mr Powell 2009 Index