Introduction In this chapter you will learn about

Introduction • In this chapter you will learn about calculations involving work, energy and power • You will learn how to use several formulae • You will learn how to solve problems involving kinetic and potential energy • You will also learn about the work-energy principle which is a common type of question on exam papers!

Work, energy and power You can calculate the work done by a force when its point of application moves by using the following formula A box is pulled 7 m across a horizontal floor by a horizontal force of magnitude 15 N. Calculate the work done by the force W = work done F = magnitude of the force s = the distance moved in the direction of the force For work done against gravity: W = work done m = mass of the object g = gravitational constant h = the height raised 15 N Sub in values from the question Calculate Work done is measured in Joules! These two formulae are effectively the same! 3 A

Work, energy and power You can calculate the work done by a force when its point of application moves by using the following formula A packing case is pulled across a horizontal floor by a horizontal rope. The case moves at a constant speed and there is a constant resistance to motion of magnitude R Newtons. When the case has moved a distance of 12 m the work done is 96 J. Calculate the magnitude of the resistance In this case you will need to use more than one formula, as we do not know either the force or the resistance… Draw a diagram – we do not know the force or the resistance, and the acceleration is 0 (constant speed) 0 RN 8 N FN Find the force acting on the box by using one of the formulae above… Sub in values from the question Calculate Now use F = ma, resolving horizontally Acceleration is 0, remember to include forces correctly Calculate 3 A

Work, energy and power Draw a diagram – Tension is the force in the cable. The weight can be added to the diagram as well and acceleration 30 g T You can calculate the work done by a force when its point of application moves by using the following formula A bricklayer raises a load of bricks of total mass 30 kg at a constant speed by attaching a cable to the bricks. 0 Assuming the cable is vertical, calculate the work done when the bricks are raised a distance of 7 m If we are going to calculate the work done, we need the tension Use F = ma and resolve vertically 30 g Calculate Rearrange Sub in values (you could also have used W = mgh) Calculate the value in Joules 3 A

Work, energy and power You can calculate the work done by a force when its point of application moves by using the following formula Draw a diagram and label all the forces R Diagram of the distance moved P 12 m A package of mass 2 kg is pulled at a constant speed up a rough plane which is inclined at an angle of 30° to the horizontal. The coefficient of friction between the package and the surface is 0. 35. The package is pulled 12 m up a line of greatest slope of the plane. Calculate: a) The work done against gravity FMAX 30° 2 g. Cos 30 30° 12 Sin 30 12 Cos 30 2 g. Sin 30 To calculate the work done against gravity, we need to know the change in vertical height of the package You can draw a diagram to show this, with the diagonal being 12 m, and the inclination still being 30° Sub values in b) The work done against friction Calculate 3 A

Work, energy and power You can calculate the work done by a force when its point of application moves by using the following formula Draw a diagram and label all the forces R P Diagram of the distance moved 12 m A package of mass 2 kg is pulled at a constant speed up a rough plane which is inclined at an angle of 30° to the horizontal. The coefficient of friction between the package and the surface is 0. 35. The package is pulled 12 m up a line of greatest slope of the plane. Calculate: a) The work done against gravity FMAX 30° 2 g. Cos 30 30° 12 Sin 30 12 Cos 30 2 g. Sin 30 We can calculate the work done against friction by using the formula W = Fs F = the force in the opposite direction to friction (as the work is done AGAINST friction) s = the distance travelled up the plane b) The work done against friction Find the force acting against FMAX We therefore need to find FMAX first, and can then use it to find the pulling force P, which is acting against friction… 3 A

Work, energy and power You can calculate the work done by a force when its point of application moves by using the following formula Draw a diagram and label all the forces 2 g. Cos 30 R P Diagram of the distance moved 12 m A package of mass 2 kg is pulled at a constant speed up a rough plane which is inclined at an angle of 30° to the horizontal. The coefficient of friction between the package and the surface is 0. 35. The package is pulled 12 m up a line of greatest slope of the plane. Calculate: a) The work done against gravity b) The work done against friction FMAX 0. 7 g. Cos 30 30° 2 g. Cos 30 30° 12 Sin 30 12 Cos 30 2 g. Sin 30 The normal reaction will just be 2 g. Cos 30 as there is no acceleration perpendicular to the plane Sub in values Simplify (to ensure it stays exact) Find the force acting against FMAX 3 A

Work, energy and power Draw a diagram and label all the forces You can calculate the work done by a force when its point of application moves by using the following formula 2 g. Cos 30 0. 7 g. Cos 30 P + 2 g. Sin 30 Diagram of the distance moved 12 m A package of mass 2 kg is pulled at a constant speed up a rough plane which is inclined at an angle of 30° to the horizontal. The coefficient of friction between the package and the surface is 0. 35. The package is pulled 12 m up a line of greatest slope of the plane. Calculate: a) The work done against gravity b) The work done against friction In these types of questions, the work done against friction and the work done against gravity give the total work done… 0. 7 g. Cos 30 30° 2 g. Cos 30 30° 12 Sin 30 12 Cos 30 2 g. Sin 30 Now resolve parallel to the plane to find force P Sub in values, acceleration is 0. remember to include the gravitational part (for now…) Work out P and leave as an exact answer Sub in F(P) and s Calculate – this gives us the TOTAL work done on the particle Subtract the work done against gravity (118 J) to leave the work done against friction 3 A

Work, energy and power 27 N You can calculate the work done by a force when its point of application moves by using the following formula A sledge is pulled 15 m across a smooth sheet of ice by a force of magnitude 27 N. The force is inclined at 25° to the horizontal. By modelling the sledge as a particle, calculate the work done by the force. As the force is at an angle to the motion, you must split it into its component parts The force will act vertically and horizontally 25° 27 Sin 25 27 Cos 25 Sub in values Calculate The total work done is 367 J However, as there is no distance travelled vertically (s = 0), there is no work done in this direction Therefore, you only need the work done horizontally… 3 A

Work, energy and power You can calculate the kinetic energy of a moving particle, and the potential energy of a particle above ground level Kinetic Energy Kinetic energy is the energy a body possesses due to its motion Faster movement = more Kinetic Energy Heavier object = more Kinetic Energy m is the mass of the particle v is its velocity Potential Energy m is the mass of the particle g is the gravitational constant h is the height of the particle above the ground (or a given fixed point) Potential energy is energy which is effectively stored in an object and which could become active A ball held in the air has potential energy, which will become kinetic energy if the ball is dropped Heavier object = more potential energy Object held higher up = more potential energy (This chapter focuses on gravitational potential energy) 3 B

Work, energy and power You can calculate the kinetic energy of a moving particle, and the potential energy of a particle above ground level Replace a with the expression we worked out The work done by a force which accelerates a particle is connected to the kinetic energy of the particle Multiply top by m Work done = Change in kinetic energy Multiply all by s To show this, we will rewrite one of the SUVAT equations to give it in terms of a. Rewrite right side Subtract u 2 Divide by 2 s Fs = work done Final KE - Initial KE 3 B

Work, energy and power You can calculate the kinetic energy of a moving particle, and the potential energy of a particle above ground level A particle of mass 0. 3 kg is moving at a speed of 9 ms-1. Calculate its kinetic energy. Sub in values Calculate 3 B

Work, energy and power You can calculate the kinetic energy of a moving particle, and the potential energy of a particle above ground level A box of mass 1. 5 kg is pulled across a smooth horizontal surface by a horizontal force. The initial speed of the box is ums-1 and its final speed is 3 ms-1. The work done by the force is 1. 8 J. Calculate the value of u. We know W, v and m, and need u Sub in values Work out parts Rearrange Divide by 0. 75 Square root Use the formula for the change in kinetic energy! 3 B

Work, energy and power You can calculate the kinetic energy of a moving particle, and the potential energy of a particle above ground level 500 N A bus of mass 2000 kg starts from rest at some traffic lights. After travelling 400 m the bus’s speed is 12 ms-1. A constant resistance of 500 N acts on the bus. Calculate the driving force, P, which can be assumed to be constant. P Replace W with Fs Sub in values We know the following pieces of information: u = 0 ms-1 v = 12 ms-1 s = 400 m m = 2000 kg We also know the overall force will be the driving force subtract the resistances F = P - 500 Calculate parts Divide by 400 Add 500 3 B

Work, energy and power You can calculate the kinetic energy of a moving particle, and the potential energy of a particle above ground level A load of bricks is lowered vertically to the ground through a distance of 15 m. Find the loss in potential energy. In this case, you can use ‘h’ as the change in height, rather than the height of the particle Sub in values. The height has fallen by 15 m… Calculate So the loss of potential energy is 4410 J 3 B

Work, energy and power You can calculate the kinetic energy of a moving particle, and the potential energy of a particle above ground level Draw a diagram 7 m A parcel of mass 3 kg is pulled 7 m up a plane inclined at an angle θ to the horizontal, where tanθ = 3/4. Assuming that the parcel moves up a line of greatest slope of the plane, calculate the potential energy gained by the parcel. You have seen situations like this before, with the angle given as Tanθ. Start by finding Sinθ and Cosθ. θ 7 Cosθ 7 Sinθ The change in potential energy will be affected by the change in the vertical height of the parcel Sub in values, using the change in height Also use the value of Sinθ Calculate 3 B

Work, energy and power You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle “The change in total energy of a particle is equal to the work done on the particle. ” (This is called the ‘work-energy’ principle) If gravity is the only force acting on a particle: “When no external forces (other than gravity) act on a particle, the sum of its potential and kinetic energies remain constant. ” (This is called the principle of the conservation of mechanical energy) If another force (usually friction) is acting on the particle: 3 C

Work, energy and power Draw a diagram The normal reaction is doing no work as there is no movement perpendicular to the plane The plane is smooth so the particle does not have to do any work against friction We can therefore use the upper of the formulae shown… A You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle B If gravity is the only force acting on a particle: 30° Initial speed = 0 If another force is acting on the particle: Final speed = 6 Find the increase in Kinetic energy A smooth plane is inclined at 30° to the horizontal. A particle of mass 0. 5 kg slides down the slope. The particle starts from rest at point A and at point B has a speed of 6 ms -1. Find the distance AB. Sub in values Calculate 3 C

Work, energy and power Draw a diagram The normal reaction is doing no work as there is no movement perpendicular to x x. Sin 30 the plane The plane is smooth so the 30° particle does not have to do x. Cos 30 any work against friction We can therefore use the upper of the formulae shown… Final speed = 6 A You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle B If gravity is the only force acting on a particle: If another force is acting on the particle: 30° Initial speed = 0 Find the decrease in Potential energy (find the change in vertical height first) A smooth plane is inclined at 30° to the horizontal. A particle of mass 0. 5 kg slides down the slope. The particle starts from rest at point A and at point B has a speed of 6 ms -1. Find the distance AB. Call the diagonal distance (the one we need to find) ‘x’ Sub in values Calculate in terms of x 3 C

Work, energy and power Draw a diagram The normal reaction is doing no work as there is no movement perpendicular to x x. Sin 30 the plane The plane is smooth so the 30° particle does not have to do x. Cos 30 any work against friction We can therefore use the upper of the formulae shown… Final speed = 6 A You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle B If gravity is the only force acting on a particle: 30° Initial speed = 0 If another force is acting on the particle: A smooth plane is inclined at 30° to the horizontal. A particle of mass 0. 5 kg slides down the slope. The particle starts from rest at point A and at point B has a speed of 6 ms -1. Find the distance AB. Sub in the values we calculated Divide by 2. 45 This could be calculated using F = ma and the SUVAT equations from M 1, however in M 2 you will usually be asked specifically to use these principles… 3 C

Work, energy and power You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle If gravity is the only force acting on a particle: 45° Initial speed = 8 Final speed = 0 If another force is acting on the particle: Draw a diagram The normal reaction is doing no work as there is no movement perpendicular to the plane As the plane is rough, the particle will have to do some work against friction. You must take this into account. You will need to use the second of the formulae to the left Find the kinetic energy lost A particle of mass 2 kg is projected with speed 8 ms-1 up a rough plane inclined at 45° to the horizontal. The coefficient of friction between the particle and the plane is 0. 4. Calculate the distance the particle travels up the plane before it comes to instantaneous rest. Sub in values Calculate 3 C

Work, energy and power 45° x. Sin 45 x You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle x. Cos 45 If gravity is the only force acting on a particle: 45° Initial speed = 8 Final speed = 0 If another force is acting on the particle: Draw a diagram The normal reaction is doing no work as there is no movement perpendicular to the plane As the plane is rough, the particle will have to do some work against friction. You must take this into account. You will need to use the second of the formulae to the left Find the potential energy gained As in the last example, call the distance moved up the plane ‘x’, and work out the vertical change, based on this… A particle of mass 2 kg is projected with speed 8 ms-1 up a rough plane inclined at 45° to the horizontal. The coefficient of friction between the particle and the plane is 0. 4. Calculate the distance the particle travels up the plane before it comes to instantaneous rest. Sub in values Calculate in terms of x 3 C

Work, energy and power x You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle If gravity is the only force acting on a particle: 45° Initial speed = 8 If another force is acting on the particle: A particle of mass 2 kg is projected with speed 8 ms-1 up a rough plane inclined at 45° to the horizontal. The coefficient of friction between the particle and the plane is 0. 4. Calculate the distance the particle travels up the plane before it comes to instantaneous rest. Final speed = 0 Draw a diagram The normal reaction is doing no work as there is no movement perpendicular to the plane As the plane is rough, the particle will have to do some work against friction. You must take this into account. You will need to use the second of the formulae to the left This time, we cannot just set these values equal to each other, as some energy will be lost to friction Find an expression for the loss of energy by using the highlighted formula Sub in values to find an expression for the loss of energy 3 C

Work, energy and power 2 g. Cos 45 R You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle FMAX 7. 84 Cos 45 If gravity is the only force acting on a particle: 45° Initial speed = 8 If another force is acting on the particle: The energy lost will all have been used against friction A particle of mass 2 kg is projected with speed 8 ms-1 up a rough plane inclined at 45° to the horizontal. The coefficient of friction between the particle and the plane is 0. 4. Calculate the distance the particle travels up the plane before it comes to instantaneous rest. Draw a diagram The normal reaction is doing x no work as there is no movement perpendicular to the plane 2 g. Cos 45 As the plane is rough, the 45° particle will have to do some 2 g work against friction. You 2 g. Sin 45 must take this into account. You will need to use the second of the formulae to Final speed = 0 the left We need to find an expression for the work done by friction, and set it equal to the loss of energy We will first need to find the normal reaction, then find the maximum frictional force Sub in values Rewrite 3 C

Work, energy and power 2 g. Cos 45 You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle 7. 84 Cos 45 If gravity is the only force acting on a particle: 45° Initial speed = 8 If another force is acting on the particle: Draw a diagram The normal reaction is doing x no work as there is no movement perpendicular to the plane 2 g. Cos 45 As the plane is rough, the 45° particle will have to do some 2 g work against friction. You 2 g. Sin 45 must take this into account. You will need to use the second of the formulae to Final speed = 0 the left Now we can calculate the work done against friction, by using one of the formulae from earlier in the chapter The frictional force acts over a distance ‘x’ A particle of mass 2 kg is projected with speed 8 ms-1 up a rough plane inclined at 45° to the horizontal. The coefficient of friction between the particle and the plane is 0. 4. Calculate the distance the particle travels up the plane before it comes to instantaneous rest. This is the energy lost Sub in F and s Rewrite in terms of x This is the work done against friction These expressions will be equal as all the energy lost has been working against friction! 3 C

Work, energy and power 2 g. Cos 45 You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle 7. 84 Cos 45 If gravity is the only force acting on a particle: 45° Initial speed = 8 If another force is acting on the particle: Set the two expressions equal to each other and solve for the distance, ‘x’ A particle of mass 2 kg is projected with speed 8 ms-1 up a rough plane inclined at 45° to the horizontal. The coefficient of friction between the particle and the plane is 0. 4. Calculate the distance the particle travels up the plane before it comes to instantaneous rest. Draw a diagram The normal reaction is doing x no work as there is no movement perpendicular to the plane 2 g. Cos 45 As the plane is rough, the 45° particle will have to do some 2 g work against friction. You 2 g. Sin 45 must take this into account. You will need to use the second of the formulae to Final speed = 0 the left Add 9. 8√ 2 x Factorise right side Divide by the bracket Calculate 3 C

Work, energy and power A You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle B 50 m 25 m If gravity is the only force acting on a particle: Initial speed = 6 The work done by the skier will be equal to the work done against resistances, subtract the total loss of energy If another force is acting on the particle: Final speed = 4 Calculate the loss of kinetic energy (it is a loss as speed has fallen) A skier passes a point A on a ski-run, moving downhill at 6 ms-1. After descending 50 m vertically, the run starts to ascend. When the skier has ascended 25 m to point B her speed is 4 ms-1. The skier and skis have a combined mass of 55 kg. The total distance travelled from A to B is 1400 m. The resistances to motion are constant and have a magnitude of 12 N. Sub in values Calculate the work done by the skier 3 C

Work, energy and power A You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle B 50 m 25 m If gravity is the only force acting on a particle: Initial speed = 6 Final speed = 4 If another force is acting on the particle: The work done by the skier will be equal to the work done against resistances, subtract the total loss of energy Calculate the gain of potential energy (it is actually a loss as the height has fallen!) A skier passes a point A on a ski-run, moving downhill at 6 ms-1. After descending 50 m vertically, the run starts to ascend. When the skier has ascended 25 m to point B her speed is 4 ms-1. The skier and skis have a combined mass of 55 kg. The total distance travelled from A to B is 1400 m. The resistances to motion are constant and have a magnitude of 12 N. Sub in values Calculate the work done by the skier 3 C

Work, energy and power A You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle If gravity is the only force acting on a particle: If another force is acting on the particle: 50 m 25 m Initial speed = 6 Final speed = 4 The work done by the skier will be equal to the work done against resistances, subtract the total loss of energy Calculate the total loss of energy A skier passes a point A on a ski-run, moving downhill at 6 ms-1. After descending 50 m vertically, the run starts to ascend. When the skier has ascended 25 m to point B her speed is 4 ms-1. The skier and skis have a combined mass of 55 kg. The total distance travelled from A to B is 1400 m. The resistances to motion are constant and have a magnitude of 12 N. Calculate the work done by the skier B Sub in values Calculate It makes sense that these are added together, as we have lost both Kinetic and Potential energies! 3 C

Work, energy and power A You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle If gravity is the only force acting on a particle: If another force is acting on the particle: 50 m 25 m Initial speed = 6 Final speed = 4 The work done by the skier will be equal to the work done against resistances, subtract the total loss of energy Calculate the total work done against resistances A skier passes a point A on a ski-run, moving downhill at 6 ms-1. After descending 50 m vertically, the run starts to ascend. When the skier has ascended 25 m to point B her speed is 4 ms-1. The skier and skis have a combined mass of 55 kg. The total distance travelled from A to B is 1400 m. The resistances to motion are constant and have a magnitude of 12 N. Calculate the work done by the skier B Sub in values – the resistances of 12 N act over 1400 m Calculate 16800 J of energy has been used against the resistances. The loss of kinetic and potential energy of 14025 J has contributed to this The rest will be work done by the skier 3 C

Work, energy and power You can calculate the power developed by an engine and solve problems about moving vehicles Power is the rate of doing work It is measured in Watts (W), where 1 watt = 1 joule per second Often an engine’s power will be given in kilowatts (k. W) where 1 k. W = 1000 W The power developed by an engine is given by the following formula: P = power (W) F = the driving force of the engine (N) v = velocity (ms-1) 3 D

Work, energy and power You can calculate the power developed by an engine and solve problems about moving vehicles A truck is being pulled up a slope at a constant speed of 8 ms-1 by a force of magnitude 2000 N acting parallel to the direction of motion of the truck. Calculate the power developed in kilowatts. Sub in values Calculate Change to kilowatts 3 D

Work, energy and power Draw a diagram and show forces You can calculate the power developed by an engine and solve problems about moving vehicles 600 N A car of mass 1250 kg is travelling along a horizontal road. The car’s engine is working at 24 k. W. The resistance to motion is constant and has magnitude 600 N. Calculate: v=6 a) The acceleration of the car when it is travelling at 6 ms-1 Sub in values Divide by 6 b) The maximum speed of the car We can calculate the driving force from the information given T is often used as the ‘tractive’ force of the engine P = 24000 W To calculate the acceleration we can use the formula F = ma. However, we do not know the driving force from the engine yet. T 4000 N Resolve horizontally and sub in values Calculate a At a velocity of 6 ms-1, the acceleration is 2. 72 ms-2 3 D

Work, energy and power You can calculate the power developed by an engine and solve problems about moving vehicles A car of mass 1250 kg is travelling along a horizontal road. The car’s engine is working at 24 k. W. The resistance to motion is constant and has magnitude 600 N. Calculate: a) The acceleration of the car when it is travelling at 6 ms-1 Draw a diagram and show forces 600 N 4000 N Sub in values Calculate v So the maximum speed of the car is 40 ms-1 b) The maximum speed of the car Important points to note: When the car is at its maximum speed, the resultant force will be 0 As the velocity of the car increases, the driving force falls (it is harder for a car to accelerate more if it is already at a high speed) The driving force must be 600 N! We can use this to calculate the velocity at this point… This is the maximum speed for the given power level. It is possible to increase the power in an engine (for example by changing gear), and hence the top speed will increase 3 D

Work, energy and power N T 1600 N You can calculate the power developed by an engine and solve problems about moving vehicles A car of mass 1100 kg is travelling at a constant speed of 15 ms-1 along a straight road which is inclined at 7˚ to the horizontal. The engine is working at a rate of 24 k. W. a) Calculate the magnitude of the nongravitational resistances to motion The rate of working of the engine is now increased to 28 k. W. Assuming the resistances to motion are unchanged: R 7˚ 1100 g. Cos 7 1100 g. Sin 7 As the speed is constant, the driving force must be equal to the forces opposing motion (gravity and anything else) Find the driving force first Sub in values Calculate F b) Calculate the initial acceleration of the car 3 D

Work, energy and power N 1600 N You can calculate the power developed by an engine and solve problems about moving vehicles 268. 2 N R 1100 g 7˚ 1100 g. Cos 7 A car of mass 1100 kg is travelling at a constant speed of 15 ms-1 along a straight road which is inclined at 7˚ to the horizontal. The engine is working at a rate of 24 k. W. As the speed is constant, the driving force must be equal to the forces opposing motion (gravity and anything else) a) Calculate the magnitude of the nongravitational resistances to motion Now you have the driving force, resolve parallel to the plane The rate of working of the engine is now increased to 28 k. W. Assuming the resistances to motion are unchanged: b) Calculate the initial acceleration of the car 7˚ 1100 g. Sin 7 Sub in values. Remember acceleration is 0 Rearrange to find R Calculate 3 D

Work, energy and power N 1600 N 1867 N You can calculate the power developed by an engine and solve problems about moving vehicles 268. 2 N 1100 g 7˚ 1100 g. Cos 7 A car of mass 1100 kg is travelling at a constant speed of 15 ms-1 along a straight road which is inclined at 7˚ to the horizontal. The engine is working at a rate of 24 k. W. As the speed is constant, the driving force must be equal to the forces opposing motion (gravity and anything else) a) Calculate the magnitude of the nongravitational resistances to motion For part b), we need to start again by calculating the tractive force of the vehicle The rate of working of the engine is now increased to 28 k. W. Assuming the resistances to motion are unchanged: 7˚ 1100 g. Sin 7 Sub in values Calculate (remember to use the exact value later on) b) Calculate the initial acceleration of the car 3 D

Work, energy and power N 1867 N You can calculate the power developed by an engine and solve problems about moving vehicles 268. 2 N 1100 g 7˚ 1100 g. Cos 7 A car of mass 1100 kg is travelling at a constant speed of 15 ms-1 along a straight road which is inclined at 7˚ to the horizontal. The engine is working at a rate of 24 k. W. As the speed is constant, the driving force must be equal to the forces opposing motion (gravity and anything else) a) Calculate the magnitude of the nongravitational resistances to motion Now use the F = ma formula again with the updated information… 7˚ The rate of working of the engine is now increased to 28 k. W. Assuming the resistances to motion are unchanged: b) Calculate the initial acceleration of the car 1100 g. Sin 7 Sub in values Divide by 1100 The initial acceleration will be 0. 242 ms-2 3 D

Summary • We have learnt how to solve problems involving work done on or by a particle • We have seen how to calculate Kinetic and Potential energies, and how these link together • We have seen how to perform calculations involving the power of an engine and its driving force