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© Boardworks Ltd 2006 1 of 46 These icons indicate that teacher’s 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. © Boardworks Ltd 2006 1 of 46 AS-Level Maths: Mechanics 2 for Edexcel M2.4 Work and Energy

© Boardworks Ltd 2006 2 of 46 Contents © Boardworks Ltd 2006 2 of 46 Energy Work Conservation of energy Work-energy and dissipative forces Power Examination-style questions

© Boardworks Ltd 2006 3 of 46 Mechanical 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 is 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 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 acceleration due to gravity and h is the vertical distance travelled in metres. ½ mv 2 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 500 ms -1 a)K.E., in joules, = ½ × m × v 2 = ½ × 0.5 × 8 2 = 16 b)K.E., in joules, = ½ × m × v 2 = ½ × 900 × 15 2 = 101 250 c)K.E., in joules, = ½ × m × v 2 = ½ × 0.02 × 500 2 = 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: 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 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. = ½ × m × v 2 = ½ × 800 × 10 2 = 40 000 Final K.E. = ½ × m × v 2 = ½ × 800 × 25 2 = 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 the 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 30 g 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 Conservation 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 force. If the direction of the force is different to the direction of motion then  F Direction of motion Work = F × d cos  = Fd cos 

© 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 × distance = 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 = Fd cos  = 8 × 3 × cos60° = 12 Therefore work done by the force is 12 J. 60 o 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. 0.05 g N Work done = force × distance = (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 is 0.3. 60° R F 5 g 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 = 5 g cos60° = 2.5 g F = µR = 0.3 × 2.5 g = 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 calculate the vertical distance travelled.  Work done against gravity = 5 g × 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 Examination-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 particle 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 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° 5g5g Since the only force acting on the particle is gravity, energy is conserved. Initial K.E. = 0 (the particle is at rest) Final K.E. = ½ × 5 × 4.9 2 = 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° 

© Boardworks 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, find 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.

© Boardworks Ltd 2006 22 of 46 Conservation question 2 45° 2g2g 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. = ½ × 2 × v 2 Final K.E. = 0 (particle comes to rest) ½ × 2 × v 2 = 69.3  v = 8.32 (3 s.f.) Therefore the particle is projected up the 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 Work-energy principle The relationship between work and energy is a simple one: This can be seen from the following application of Newton’s 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 speed 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 Newton’s Second Law: F = ma  a = F / m This gives Fd = ½ mv 2 – ½ mu 2 Since the force is assumed constant, the acceleration can also be assumed constant. Using v 2 = u 2 + 2 as v 2 = u 2 + 2 × F / m × d v 2 = u 2 +  ½ mv 2 = ½ mu 2 + 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 when it has travelled 6 m. 10 N 3 kg Direction of motion Work done = Fd = 10 × 6 = 60 Fd = ½ mv 2 – ½ mu 2 Work done = ½ mv 2 – ½ mu 2 = 60 ½ × 3 × v 2 – 0 = 60 1.5 v 2 = 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: 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. = ½ × 2.5 × 2 2 = 5 Final K.E. = ½ × 2.5 × 5 2 = 31.25  Change in kinetic energy = 26.25 Work done = Fd = 10 F = 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. Friction 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 rough 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. = ½ × 3 × 4 2 – ½ × 3 × 3 2 = 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 done 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 doing 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 by:

© 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 = = = 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 the 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 = 18 000 ÷ 20 = 900 N c) Force produced = 18 000 ÷ 30 = 600 N Power = Force × VelocitySo,

© 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 of 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 questions 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 and 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 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 Newton’s Second Law, F = ma 3000 – 500 = 1000 a 1000 a = 2500  a = 2.5 Therefore the acceleration of the car when it is travelling at 15 ms -1 is 2.5 ms -2. 500 N D 1000 a a) To calculate the acceleration it is first necessary to calculate the driving force. So,

© Boardworks Ltd 2006 39 of 46 Exam question 1 R D 1000 g 500 N  Applying Newton’s Second Law up the slope, D – 500 – 1000 g sin  = 0 D = 500 + 1000 g × 0.1 = 1480 The driving force of 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.

© Boardworks Ltd 2006 40 of 46 Exam question 2 Examination-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) Calculate 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. = ½ × 0.5 × 10 2 Final K.E. = ½ × 0.5 × 6 2 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.f.) 16 + 12.7 = 28.7  h = 10sin15° Therefore the loss of energy of the stone is 28.7 J.

© Boardworks Ltd 2006 42 of 46 Exam question 2 R = 0.5 g cos15° = 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 against friction. The work done against friction over a distance of 10 m is 28.7 J. Work done = force × distance  force = 28.7 ÷ 10 = 2.87 R F 0.5 g 15°

© Boardworks Ltd 2006 43 of 46 Exam question 3 Examination-style question 3: A parcel of mass 30 kg is hauled up a slope inclined at an angle of 20° to the horizontal. The parcel is hauled by means of a constant force of 300 N at an angle of 30° to the line of greatest slope of the plane. The acceleration of the parcel up the plane is 0.1 ms -2 and the total resistance to motion is R N. a) Find the work done by the force in hauling the parcel, from rest, 10 m up the slope. b) By considering work and energy, find the value of R.

© Boardworks Ltd 2006 44 of 46 Exam question 3 a) Work done = force × distance Since the force acts at an angle of 30° to the slope, the component of force acting along the slope must be calculated. 300 N 30° c  c = 300cos30 o = 260 (3 s.f) Work done = 260 × 10 = 2600 So, work done by the force is 2600 J.

© Boardworks Ltd 2006 45 of 46 Exam question 3 10 20° h b) so, h = 10sin20° To find gain in K.E. it is necessary to find the speed of the parcel after it has travelled 10 m. Gain in K.E. = ½ × 30 × 2 = 30 Gain in G.P.E. = 30 × 9.8 × 10sin20 o = 1006 (4 s.f) v 2 = u 2 + 2 as = 0 + 2 × 0.1 × 10 = 2 v = 1.41 ms -1 (3 s.f.)

© Boardworks Ltd 2006 46 of 46 Exam question 3 The difference between the work done by the force acting up the slope and the gain in energy of the system is 2600 – 1036 = 1564 J. This is the work done by the resistance force. Work done = force × distance  R = 1564 ÷ 10 = 156.4 N Therefore the increase in energy of the system is 1036 J (4 s.f). 1006 + 30 = 1036

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