M2.6 Energy, Work and Power

1、 boardworks ltd 2006 1 of 46 these icons indicate that teachers notes or useful web addresses are available in the notes page. this icon indicates the slide contains activities created in flash. these activities are not editable. for more detailed instructions, see the getting started presentation.

2、boardworks ltd 2006 1 of 46 as-level maths: mechanics 2 for ocr m2.6 energy, work and power boardworks ltd 2006 2 of 46 contents boardworks ltd 2006 2 of 46 energy energy work conservation of energy work-energy and dissipative forces power examination-style questions boardworks ltd 2006 3 of 46 mech

3、anical energy energy is measured in joules or kilojoules, represented by j or kj. mechanical energy is energy that a particle has due to either its motion or its position. energy exists in many forms. these include electrical energy, chemical energy, heat energy and light energy. mechanical energy i

4、s the energy that will be considered in m2. it comes in two forms kinetic and potential energy. energy is a measure of a particles capacity to do work. boardworks ltd 2006 4 of 46 kinetic and potential energy kinetic energy kinetic energy is the energy a particle possesses due to its motion. kinetic

5、 energy is defined as where m is mass in kg and v is speed in ms-1. gravitational potential energy gravitational potential energy is the energy a particle possesses due to its position. it is stored energy. change in gravitational potential energy is defined as where m is mass in kg, g is accelerati

6、on due to gravity and h is the vertical distance travelled in metres. mv2 mgh, boardworks ltd 2006 5 of 46 kinetic energy find the kinetic energy of the following: a) a particle of mass 0.5 kg travelling at 8 ms-1 b) a car of mass 900 kg travelling at 15 ms-1 c) a bullet of mass 20 g travelling at 5

7、00 ms-1 a) k.e., in joules, = m v2 = 0.5 82 = 16 b) k.e., in joules, = m v2 = 900 152 = 101 250 c) k.e., in joules, = m v2 = 0.02 5002 = 2500 boardworks ltd 2006 6 of 46 find the gravitational potential energy of the following, stating whether it is a gain or loss in gravitational potential energy:

8、gravitational potential energy a) g.p.e., in joules, = m g h = 2 9.8 8 = 156.8this is a gain in g.p.e. b) g.p.e., in joules, = m g h = 500 9.8 50 = 245 000this is a loss of g.p.e. c) g.p.e., in joules, = m g h = 80 9.8 75 = 58 800this is a gain in g.p.e. a) a particle of mass 2 kg raised a vertical

9、distance of 8 m b) a lift of mass 500 kg descending 50 m vertically c) a man of mass 80 kg climbing a vertical distance of 75 m. boardworks ltd 2006 7 of 46 energy question 1 question 1: find the gain in kinetic energy of a car of mass 800 kg as it accelerates from 10 ms-1 to 25 ms-1. initial k.e. =

10、 m v2 = 800 102 = 40 000 final k.e. = m v2 = 800 252 = 250 000 therefore the gain in kinetic energy of the car is 210 000 j. boardworks ltd 2006 8 of 46 energy question 2 question 2: a child of mass 30 kg slides down a slide of length 4 m inclined at an angle of 30 to the horizontal. by modelling th

11、e child as a particle and the slide as a smooth inclined plane, find the loss of gravitational potential energy of the child. 30 4 m 30g n g.p.e. = m g h = 30 9.8 4sin30 = 588 therefore the loss in g.p.e. is 588 j. boardworks ltd 2006 9 of 46 contents boardworks ltd 2006 9 of 46 work energy work con

12、servation of energy work-energy and dissipative forces power examination-style questions boardworks ltd 2006 10 of 46 work work can be defined as force distance i.e., the work done by a constant force is the force (in newtons) multiplied by the distance moved (in metres) in the direction of the forc

13、e. if the direction of the force is different to the direction of motion then f direction of motion work = f dcos = fdcos boardworks ltd 2006 11 of 46 work question 1 question 1: find the work done by a force of 10 n acting on a particle that moves 3 m in the direction of the force. work = force dis

14、tance = 10 3 = 30 therefore the work done by the force is 30 j. boardworks ltd 2006 12 of 46 work question 2 question 2: a particle of mass 2 kg is pulled 3 m along a smooth horizontal surface by a force of 8 n acting at an angle of 60 to the horizontal. find the work done by the force. work done =

15、fdcos = 8 3 cos60 = 12 therefore work done by the force is 12 j. 60o 8 n direction of motion 2 kg boardworks ltd 2006 13 of 46 work question 3 question 3: a stone of mass 50 g falls from the top of a cliff of height 75 m. find the work done by gravity as the stone falls. 0.05g n work done = force di

16、stance = (0.05 9.8) 75 = 36.75 therefore the work done by gravity is 36.75 j. boardworks ltd 2006 14 of 46 work question 4 question 4: a particle of mass 5 kg is pulled 8 m up a rough plane inclined at an angle of 60 to the horizontal. the coefficient of friction between the plane and the particle i

17、s 0.3. 60 r f 5g n a) find the work done against friction. b) find the work done against gravity. boardworks ltd 2006 15 of 46 work question 4 a) to find the work done against the frictional force it is first necessary to find the frictional force. resolving perpendicular to the plane: r = 5gcos60 =

18、 2.5g f = r = 0.3 2.5g = 7.35 the frictional force is 7.35 n. the distance travelled in the direction opposite to the frictional force is 8 m. work done against friction = 7.35 8 = 58.8 j. boardworks ltd 2006 16 of 46 work question 4 b) to find the work done against gravity it is necessary to calcul

19、ate the vertical distance travelled. work done against gravity = 5g 8sin60 = 339 j (3 s.f.) 8 m h 60 sin60 = h 8 h = 8sin60 boardworks ltd 2006 17 of 46 contents boardworks ltd 2006 17 of 46 conservation of energy energy work conservation of energy work-energy and dissipative forces power examinatio

20、n-style questions boardworks ltd 2006 18 of 46 conservation of energy the total mechanical energy of a particle remains constant if the only force acting on the particle is gravity. the total mechanical energy of a particle is the sum of any kinetic energy and any potential energy. hence, for a part

21、icle on which the only force acting is gravity, initial k.e. + initial p.e. = final k.e. + final p.e. therefore loss in k.e. = gain in p.e. and gain in k.e. = loss in p.e. boardworks ltd 2006 19 of 46 conservation question 1 question 1: a particle of mass 5 kg slides down a smooth slope inclined at

22、an angle of 30 to the horizontal. given that the particle starts from rest, find the distance travelled along the slope when the particle has reached a speed of 4.9 ms-1. 30 5g since the only force acting on the particle is gravity, energy is conserved. initial k.e. = 0 (the particle is at rest) fin

23、al k.e. = 5 4.92 = 60.025 gain in k.e. = 60.025 j loss in g.p.e. = 60.025 j boardworks ltd 2006 20 of 46 conservation question 1 change in g.p.e. = 5 9.8 h = 60.025 therefore the distance moved along the slope is 2.45 m. d 1.225 m 30 . . 60 025 1 225 m 49 h .1 225 sin30 = d . . 1 225 2 45 sin30 d bo

24、ardworks ltd 2006 21 of 46 conservation question 2 question 2: a particle of mass 2 kg is projected up a rough plane inclined at an angle of 45 to the horizontal. the coefficient of friction between the particle and the plane is 0.15. if the particle travels 5 m up the plane before coming to rest, f

25、ind the speed with which the particle was projected. since the only force acting on the particle is gravity, energy is conserved. h = 5sin45 therefore the vertical distance travelled is 3.54 m (3 s.f.). 5 m h 45 to calculate the gain in g.p.e. we need to calculate the vertical height gained. sin45 =

26、 5 h boardworks ltd 2006 22 of 46 conservation question 2 45 2g gain in g.p.e. = mgh = 2 9.8 5sin45 = 69.3 (3 s.f.) as energy is conserved, gain in g.p.e. = loss in k.e. initial k.e. = 2 v2 final k.e. = 0 (particle comes to rest) 2 v2 = 69.3 v = 8.32 (3 s.f.) therefore the particle is projected up t

27、he plane with a speed of 8.32 ms-1. loss in k.e. gain in g.p.e. boardworks ltd 2006 23 of 46 contents boardworks ltd 2006 23 of 46 work-energy and dissipative forces energy work conservation of energy work-energy and dissipative forces power examination-style questions boardworks ltd 2006 24 of 46 w

28、ork-energy principle the relationship between work and energy is a simple one: this can be seen from the following application of newtons second law. a particle of mass m kg is moved a distance of d m horizontally by a constant horizontal force f n. let the initial speed be u ms-1 and the final spee

29、d be v ms-1. work done = change in kinetic energy boardworks ltd 2006 25 of 46 remember: s = distance u = initial velocity v = final velocity work-energy principle by newtons second law: f = ma a = f/m this gives fd = mv2 mu2 since the force is assumed constant, the acceleration can also be assumed

30、constant. using v2 = u2 + 2as v2 = u2 + 2 f/m d 2fd m v2 = u2 + mv2 = mu2 + fd boardworks ltd 2006 26 of 46 work-energy question 1 question 1: a particle of mass 3 kg is pulled 6 m by a horizontal force of 10 n across a smooth horizontal surface. if the particle started from rest, find its speed whe

31、n it has travelled 6 m. 10 n 3 kg direction of motion work done = fd = 10 6 = 60 fd = mv2 mu2 work done = mv2 mu2 = 60 3 v2 0 = 60 1.5v2 = 60 v = 6.32 (3 s.f.) therefore the speed of the particle when it has travelled 6 m is 6.32 ms-1. boardworks ltd 2006 27 of 46 work-energy question 2 question 2:

32、find the horizontal force that causes a particle of mass 2.5 kg to increase speed from 2 ms-1 to 5 ms-1 over a distance of 10 m on a smooth horizontal surface. work done = change in kinetic energy initial k.e. = 2.5 22 = 5 final k.e. = 2.5 52 = 31.25 change in kinetic energy = 26.25 work done = fd =

33、 10f = 26.25 f = 2.625 therefore the required force is 2.625 n. boardworks ltd 2006 28 of 46 work and dissipative forces work done against a dissipative force is equal to the loss of the total mechanical energy of the system. a dissipative force is a force which causes a particle to lose energy. fri

34、ction and air resistance are examples of dissipative forces. this means that the principle of conservation no longer applies, as the total mechanical energy of the particle is not constant. boardworks ltd 2006 29 of 46 work and dissipative forces example: a particle of mass 3 kg is moving down a rou

35、gh slope inclined at an angle of 30 to the horizontal. the particle passes a point a at 3 ms-1 and a point b 2 m down the slope at 4 ms-1. calculate the work done against friction. gain in k.e. = 3 42 3 32 = 10.5 j loss in g.p.e. = 3 g 1 = 29.4 j sin30 = h 2 loss of energy = 29.4 10.5 = 18.9 work do

36、ne against friction = 18.9 j h 2 m 30 h = 2sin30 = 1 boardworks ltd 2006 30 of 46 contents boardworks ltd 2006 30 of 46 power energy work conservation of energy work-energy and dissipative forces power examination-style questions boardworks ltd 2006 31 of 46 power power is defined as the rate of doi

37、ng work, with respect to time. the s.i. unit of power is the watt, which is a rate of work of 1 joule per second. the watt is a relatively small unit. you will more commonly meet the larger unit, the kilowatt. 1 kilowatt (abbreviated to 1 kw) = 1000 watts = 1000 joules per second power is calculated

38、 by: work (in joules) time (in seconds) boardworks ltd 2006 32 of 46 power consider a force f acting for a small time interval t over a small distance d. work done = f d so, power = so, the power of an engine or machine moving at a speed v subject to a driving force f is fv. power = work (in joules)

39、 time (in seconds) fd t = d f t = fv boardworks ltd 2006 33 of 46 power question 1 question 1: a car exerts a driving force of 1500 n. if the car is travelling at a constant speed of 35 ms-1, calculate the power of the engine. power, in watts = fv = 1500 35 = 52 500 therefore the power produced by t

40、he car engine is 52.5 kw. boardworks ltd 2006 34 of 46 power question 2 question 2: the engine of a car is generating power of 18 kw. find the driving force produced when the car is travelling at a speed of: a) 12 ms-1 b) 20 ms-1 c) 30 ms-1 a) force produced = 18 000 12 = 1500 n b) force produced =

41、18 000 20 = 900 n c) force produced = 18 000 30 = 600 n power = force velocity power force = velocity so, boardworks ltd 2006 35 of 46 power question 3 question 3: a cyclist is travelling along a horizontal road at a constant speed of 7 ms-1. if the cyclist experiences a total resistance to motion o

42、f 20 n, find the power produced by the cyclist. since the speed is constant there is no acceleration. therefore, driving force = resistance force. driving force = 20 n velocity = 7 ms-1 power = 20 7 = 140 w boardworks ltd 2006 36 of 46 contents boardworks ltd 2006 36 of 46 examination-style question

43、s energy work conservation of energy work-energy and dissipative forces power examination-style questions boardworks ltd 2006 37 of 46 exam question 1 examination-style question 1: a car of mass 1000 kg moves along a straight, horizontal road. the car engine is working at a constant rate of 45 kw an

44、d the total resistance to the motion of the car is 500 n. a) find the acceleration of the car when its speed is 15 ms-1. the car comes to a hill which is inclined at an angle of to the horizontal, where sin = 0.1. the resistance to the motion of the car is unchanged. b) find the maximum speed of the

45、 car up the hill when it is working at a rate of 45 kw. boardworks ltd 2006 38 of 46 exam question 1 power = force velocity applying newtons second law, f = ma 3000 500 = 1000a 1000a = 2500 a = 2.5 therefore the acceleration of the car when it is travelling at 15 ms-1 is 2.5 ms-2. 500 nd 1000 a a) t

46、o calculate the acceleration it is first necessary to calculate the driving force. power force = velocity so, 45 000 = = 3000 15 d boardworks ltd 2006 39 of 46 exam question 1 r d 1000g 500 n applying newtons second law up the slope, d 500 1000gsin = 0 d = 500 + 1000g 0.1 = 1480 the driving force of

47、 the car is 1480 n. power = force velocity b) before calculating the maximum speed of the car it is first necessary to calculate the driving force. the maximum speed of the car is 30.4 ms-1. power velocity = force .() 45 000 30 4 3 s.f. 1480 v boardworks ltd 2006 40 of 46 exam question 2 examination

48、-style question 2: a stone of mass 0.5 kg is sliding down a rough plane inclined at an angle of 15 to the horizontal. the stone passes a point a with a speed of 10 ms-1 and a point b with a speed of 6 ms-1. a) find the loss of mechanical energy of the stone as it moves from point a to point b. b) ca

49、lculate the coefficient of friction between the stone and the plane. boardworks ltd 2006 41 of 46 exam question 2 a) loss of energy of the stone = loss of k.e. + loss of g.p.e. initial k.e. = 0.5 102 final k.e. = 0.5 62 loss of k.e. = 25 9 = 16 j 15 10 h loss of g.p.e. = 0.5 9.8 10sin15 = 12.7 (3 s.

50、f.) 16 + 12.7 = 28.7 h = 10sin15 therefore the loss of energy of the stone is 28.7 j. sin15 10 h boardworks ltd 2006 42 of 46 exam question 2 r = 0.5gcos15 = 4.73 (3 s.f.) b) to calculate the coefficient of friction, it is necessary to calculate both the normal contact force and the frictional force. the coefficient of friction is 0.61. loss of energy of the stone found in part a) is equal to the work done again