1. 2.1d Mechanics Work, energy and power Breithaupt pages 148 to 159 April 14 th, 2012

2. AQA AS Specification LessonsTopics 1 & 2Work, energy and power W = Fs cos θ P = ΔW / Δt P = Fv 3 & 4Conservation of energy Principle of conservation of energy, applied to examples involving gravitational potential energy, kinetic energy and work done against resistive forces. ΔE p = mgΔh E k = ½ mv 2

3. Work (W) Work is done when a force moves its point of application. work = force x distance moved in the direction of the force W = F s unit: joule (J) work is a scalar quantity

4. If the direction of the force and the distance moved are not in the same direction: W = F s cos θ The point of application of force, F moves distance s cos θ when the object moves through the distance s. F s θ object

5. Question 1 Calculate the work done when a force of 5 kN moves through a distance of 30 cm work = force x distance = 5 kN x 30 cm = 5000 N x 0.30 m work = 1500 J

6. Question 2 Calculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cm. work = force x distance the child must exert an upward force equal to its weight the distance moved upwards equals (12 x 20cm) = 2.4m work = 300 N x 2.4 m work = 720 J

7. Question 3 Calculate the work done by the wind on the yacht in the situation shown below: W = F s cos θ = 800 N x 50 m x cos 30° = 40 000 x cos 30° = 40 000 x 0.8660 work = 34 600 J wind force = 800 N distance moved by yacht = 50 m 30°

8. Complete: ForceDistance Angle between F and s Work 400 N5 km0°0°2 MJ 200 μN300 m0°0°60 mJ 50 N6 m60°150 J 400 N3 m90°0 J Answers 400 N 300 m 60° 0 J * * Note: No work is done when the force and distance are perpendicular to each other.

9. Force-distance graphs The area under the curve is equal to the work done. F s force distance area = work done F s force distance area = work = ½ F s area = work found by counting squares on the graph F s force distance

10. Question Calculate the work done by the brakes of a car if the force exerted by the brakes varies over the car’s braking distance of 100 m as shown in the graph below. Work = area under graph = area A + area B = (½ x 1k x 50) + (1k x 100) = (25k) + (100k) work = 125 kJ 2 force / kN distance / m 1 50 100 area B area A

11. Energy (E) Energy is needed to move objects, to change their shape or to warm them up. Work is a measurement of the energy required to do a particular task. work done = energy change unit: joule (J)

12. Conservation of Energy The principle of the conservation of energy states that energy cannot be created or destroyed. Energy can change from one form to another. All forms of energy are scalar quantities

13. Some examples of forms of energy Kinetic energy (KE) Energy due to a body’s motion. Potential energy (PE) Energy due to a body’s position Thermal energy Energy due to a body’s temperature. Chemical energy Energy associated with chemical reactions. Nuclear energy Energy associated with nuclear reactions. Electrical energy Energy associated with electric charges. Elastic energy Energy stored in an object when it is stretched or compressed. All of the above forms of energy (and others) can ultimately be considered to be variations of kinetic or potential energy.

14. Kinetic Energy (E K ) Kinetic energy is the energy an object has because of its motion and mass. kinetic energy = ½ x mass x (speed) 2 E K = ½ m v 2 Note: v = speed NOT velocity. The direction of motion has no relevance to kinetic energy.

15. Question 1 Calculate the kinetic energy of a car of mass 800 kg moving at 6 ms -1 E K = ½ m v 2 = ½ x 800kg x (6ms -1 ) 2 = ½ x 800 x 36 = 400 x 36 kinetic energy = 14 400 J

16. Question 2 Calculate the speed of a car of mass 1200kg if its kinetic energy is 15 000J E K = ½ m v 2 15 000J = ½ x 1200kg x v 2 15 000 = 600 x v 2 15 000 ÷ 600 = v 2 25 = v 2 v =  25 speed = 5.0 ms -1

17. Question 3 Calculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms -1 if its brakes exert a constant force of 3 kN. k.e. of car = ½ m v 2 = ½ x 900kg x (20ms -1 ) 2 = ½ x 900 x 400 = 450 x 400 k.e. = 180 000 J The work done by the brakes will be equal to this kinetic energy. W = F s 180 000 J = 3 kN x s 180 000 = 3000 x s s = 180 000 / 3000 braking distance = 60 m

18. Complete: MassSpeedKinetic energy 400 g4.0 ms -1 3.2 J 3000 kg10 kms -1 60 mJ 8 kg300 cms -1 36 J 50 mg12 ms -1 3.6 mJ Answers 8 kg 12 ms -1 1.5 x 10 11 J 3.2 J

19. Gravitational Potential Energy (gpe) Gravitational potential energy is the energy an object has because of its position in a gravitational field. change in g.p.e. = mass x gravitational field strength x change in height ΔE P = m g Δh

20. Question Calculate the change in g.p.e. when a mass of 200 g is lifted upwards by 30 cm. (g = 9.8 Nkg -1 ) ΔE P = m g Δh = 200 g x 9.8 Nkg -1 x 30 cm = 0.200 kg x 9.8 Nkg -1 x 0.30 m change in g.p.e. = 0.59 J

21. Complete: massgΔhΔhΔEPΔEP 3 kg10 Nkg -1 400 cm120 J 200 g1.6 Nkg -1 30 m9.6 J 7 kg10 Nkg -1 4000 m280 kJ 2000 g24 Nkg -1 3000 mm144 J Answers 3 kg 1.6 Nkg -1 4000 m 144 J

22. Falling objects If there is no significant air resistance then the initial GPE of an object is transferred into kinetic energy. ΔE K = ΔE P ½ m v 2 = m g Δh ΔhΔh m ½ Δh v1v1 v2v2 gpe = mgΔh ke = ½ mv 2 2 ke = 0 gpe = 0 gpe = ke gpe = ½ mgΔh ke = ½ mv 1 2 ke = mgΔh

23. Question A child of mass 40 kg climbs up a wall of height 2.0 m and then steps off. Assuming no significant air resistance calculate the maximum: (a) gpe of the child (b) speed of the child g = 9.8 Nkg -1 (a) max gpe occurs when the child is on the wall gpe = mgΔh = 40 x 9.8 x 2.0 max gpe = 784 J (b) max speed occurs when the child reaches the ground ½ m v 2 = m g Δh ½ m v 2 = 784 J v 2 = (2 x 784) / 40 v 2 = 39.2 v =  39.2 max speed = 6.3 ms -1

24. Power (P) Power is the rate of transfer of energy. power = energy transfer time P = ΔE Δt unit: watt (W) power is a scalar quantity

25. Power is also the rate of doing work. power = work done time P = ΔW Δt

26. Question 1 Calculate the power of an electric motor that lifts a mass of 50 kg upwards by 3.0 m in 20 seconds. g = 9.8 Nkg -1 ΔE P = m g Δh = 50 kg x 9.8 Nkg -1 x 3 m = 1470 J P = ΔE / Δt = 1470 J / 20 s power = 74 W

27. Question 2 Calculate the power of a car engine that exerts a force of 40 kN over a distance of 20 m for 10 seconds. W = F s = 40 kN x 20 m = 40 000 x 20 m = 800 000 J P = ΔW / Δt = 800 000 J / 10 s power = 80 000 W

28. Complete: energy transfer work donetimepower 600 J 2 mins5 W 440 J 20 s22 W 28 800 J 2 hours4 W 2.5 mJ 50 μs50 W Answers 600 J5 W 440 J20 s 28 800 J 2.5 mJ 50 W

29. Power and velocity power = work done / time but: work = force x displacement therefore: power = force x displacement time but: displacement / time = velocity therefore: power = force x velocity P = F v

30. Question Calculate the power of a car that maintains a constant speed of 30 ms -1 against air resistance forces of 2 kN As the car is travelling at a constant speed the car’s engine must be exerting a force equal to the opposing air resistance forces. P = F v = 2 kN x 30 ms -1 = 2 000 N x 30 ms -1 power = 60 kW

31. Energy efficiency Energy efficiency is a measure of how usefully energy is used by a device. efficiency = useful energy transferred by the device total energy supplied to the device As the useful energy can never be greater than the energy supplied the maximum efficiency possible is 1.0

32. Also: efficiency = useful work output energy supplied useful power output efficiency = power input In all cases: percentage efficiency = efficiency x 100

33. Complete Input energy (J) Useful energy (J) Wasted energy (J) EfficiencyPercentage efficiency 10040 25050 0.20 8030% 60 200 1040 2456 120 0.80 0.50 0.30 20% 0.40 80% 50% 40% Answers

34. Internet Links Reaction time stopping a car - also plots velocity/time graph - NTNUReaction time stopping a car Car Accident & Reaction Time - NTNUCar Accident & Reaction Time Work (GCSE) - Powerpoint presentation by KTWork (GCSE) Kinetic Energy (GCSE) - Powerpoint presentation by KTKinetic Energy (GCSE) Gravitational Potential Energy (GCSE) - Powerpoint presentation by KTGravitational Potential Energy (GCSE) Energy Skate Park - Colorado - Learn about conservation of energy with a skater dude! Build tracks, ramps and jumps for the skater and view the kinetic energy, potential energy and friction as he moves. You can also take the skater to different planets or even space!Energy Skate Park Rollercoaster Demo - FunderstandingRollercoaster Demo Energy conservation with falling particles - NTNUEnergy conservation with falling particles Ball rolling up a slope- NTNUBall rolling up a slope

35. Core Notes from Breithaupt pages 148 to 159 1.What is the principle of conservation of energy? 2.Define work and give its unit. Explain how work is calculated when force and distance are not in the same direction. 3.With the aid of a diagram explain how work can be found from a graph. 4.Explain what is meant by, and give equations for (a) kinetic energy & (b) gravitational potential energy. 5.In terms of energy explain what happens as a body falls under gravity. 6.In terms of energy and work define power. 7.Show that the power of an engine is given by: P = Fv.

36. Notes from Breithaupt pages 148 to 150 Work and energy 1.What is the principle of conservation of energy? 2.Define work and give its unit. Explain how work is calculated when force and distance are not in the same direction. 3.With the aid of a diagram explain how work can be found from a graph. 4.Try the summary questions on page 150

37. Notes from Breithaupt pages 151 & 152 Kinetic and potential energy 1.Explain what is meant by, and give equations for (a) kinetic energy & (b) gravitational potential energy. 2.In terms of energy explain what happens as a body falls under gravity. 3.Repeat the worked example on page 152 this time where the track drops vertically 70 m and the train has a mass of 3000 kg. 4.Try the summary questions on page 152

38. Notes from Breithaupt pages 153 & 154 Power 1.In terms of energy and work define power. 2.Show that the power of an engine is given by: P = Fv. 3.Repeat the worked example on page 154 this time where the engine exerts a force of 50 kN with a constant velocity of 100 ms -1. 4.Try the summary questions on page 154

39. Notes from Breithaupt pages 155 & 156 Energy and efficiency 1.Try the summary questions on page 156

40. Notes from Breithaupt pages 157 to 159 Renewable energy 1.Try the summary questions on page 159