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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 7m across a horizontal floor by a horizontal force of magnitude 15N. Calculate the work done by the force 𝑾 = 𝑭𝒔 W = work done F = magnitude of the force s = the distance moved in the direction of the force For work done against gravity: 𝑾 = 𝒎𝒈𝒉 W = work done m = mass of the object g = gravitational constant h = the height raised 15N 𝑊 = 𝐹𝑠 Sub in values from the question 𝑊 = 15 × 7 𝑊 = 105𝐽 Calculate Work done is measured in Joules! These two formulae are effectively the same! 3A 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 12m the work done is 96J. 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… 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 Draw a diagram – we do not know the force or the resistance, and the acceleration is 0 (constant speed) 0 RN 8N FN Find the force acting on the box by using one of the formulae above… 𝑊 = 𝐹𝑠 Sub in values from the question 96 = 𝐹 × 12 𝐹 = 8𝑁 Calculate Now use F = ma, resolving horizontally 𝐹 = 𝑚𝑎 8−𝑅 =0 𝑅 = 8𝑁 Acceleration is 0, remember to include forces correctly Calculate 3A Work, energy and power 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 Draw a diagram – Tension is the force in the cable. The weight can be added to the diagram as well and acceleration 30g 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 30kg 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 7m If we are going to calculate the work done, we need the tension Use F = ma and resolve vertically 30g 𝐹 = 𝑚𝑎 𝑇 − 30𝑔 = 0 𝑇 = 30𝑔 𝑊 = 𝐹𝑠 𝑊 = 30𝑔 × 7 𝑊 = 2058𝐽 Calculate Rearrange Sub in values (you could also have used W = mgh) Calculate the value in Joules 3A Work, energy and power You can calculate the work done by a force when its point of application moves by using the following formula 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 Draw a diagram and label all the forces R Diagram of the distance moved P 12m A package of mass 2kg 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 12m up a line of greatest slope of the plane. Calculate: a) The work done against gravity = 118𝐽 b) The work done against friction FMAX 2g 30° 30° 2gCos30 30° 12Sin30 12Cos30 2gSin30 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 12m, and the inclination still being 30° 𝑊 = 𝑚𝑔ℎ 𝑊 = 2 × 9.8 × (12𝑆𝑖𝑛30) Sub values in Calculate 𝑊 = 118𝐽 3A Work, energy and power You can calculate the work done by a force when its point of application moves by using the following formula 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 Draw a diagram and label all the forces R P Diagram of the distance moved 12m A package of mass 2kg 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 12m up a line of greatest slope of the plane. Calculate: a) The work done against gravity = 118𝐽 b) The work done against friction Find the force acting against FMAX FMAX 30° 2g 30° 2gCos30 30° 12Sin30 12Cos30 2gSin30 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 We therefore need to find FMAX first, and can then use it to find the pulling force P, which is acting against friction… 3A Work, energy and power You can calculate the work done by a force when its point of application moves by using the following formula 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 Draw a diagram and label all the forces 2gCos30 R P Diagram of the distance moved 12m A package of mass 2kg 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 12m up a line of greatest slope of the plane. Calculate: a) The work done against gravity = 118𝐽 b) The work done against friction FMAX 0.7gCos30 2g 30° 30° 2gCos30 30° 12Sin30 12Cos30 2gSin30 The normal reaction will just be 2gCos30 as there is no acceleration perpendicular to the plane 𝐹𝑀𝐴𝑋 = 𝜇𝑅 𝐹𝑀𝐴𝑋 = 0.35 × (2𝑔𝐶𝑜𝑠30) 𝐹𝑀𝐴𝑋 = 0.7𝑔𝐶𝑜𝑠30 Sub in values Simplify (to ensure it stays exact) Find the force acting against FMAX 3A Work, energy and power You can calculate the work done by a force when its point of application moves by using the following formula 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 Draw a diagram and label all the forces 2gCos30 0.7gCos30P + 2gSin30 Diagram of the distance moved 12m A package of mass 2kg 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 12m up a line of greatest slope of the plane. Calculate: a) The work done against gravity = 118𝐽 b) The work done against friction = 70.9𝐽 0.7gCos30 2g 30° 30° 2gCos30 30° 12Sin30 12Cos30 2gSin30 Now resolve parallel to the plane to find force P 𝐹 = 𝑚𝑎 𝑃 − 0.7𝑔𝐶𝑜𝑠30 − 2𝑔𝑆𝑖𝑛30 = 0 𝑃 = 0.7𝑔𝐶𝑜𝑠30 + 2𝑔𝑆𝑖𝑛30 𝑊 = 𝐹𝑠 𝑊 = (0.7𝑔𝐶𝑜𝑠30 + 2𝑔𝑆𝑖𝑛30) × 12 In these types of questions, the 𝑊 = 188.9𝐽 work done against friction and the work done against gravity give the total work 𝑊(𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛) = 70.9𝐽 done… 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 (118J) to leave the work done against friction 3A 𝑾 = 𝑭𝒔 Work, energy and power 27N You can calculate the work done by a force when its point of application moves by using the following formula A sledge is pulled 15m across a smooth sheet of ice by a force of magnitude 27N. 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 𝑾 = 𝒎𝒈𝒉 25° 27Sin25 27Cos25 𝑊 = 𝐹𝑠 Sub in values 𝑊 = (27𝐶𝑜𝑠25) × 15 Calculate 𝑊 = 367𝐽 The total work done is 367J 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… 3A 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 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 𝐾𝐸 = 1 𝑚𝑣 2 2 m is the mass of the particle v is its velocity Potential Energy 𝑃𝐸 = 𝑚𝑔ℎ 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) Kinetic energy is the energy a body possesses due to its motion Faster movement = more Kinetic Energy Heavier object = more Kinetic Energy 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) 3B 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 Work, energy and power You can calculate the kinetic energy of a moving particle, and the potential energy of a particle above ground level The work done by a force which accelerates a particle is connected to the kinetic energy of the particle Work done = Change in kinetic energy 𝐹 = 𝑚𝑎 𝑣 2 − 𝑢2 𝐹=𝑚 2𝑠 Multiply top by m 𝑚(𝑣 2 − 𝑢2 ) 𝐹= 2𝑠 𝐹𝑠 = To show this, we will rewrite one of the SUVAT equations to give it in terms of a. 𝐹𝑠 = 𝑣 2 = 𝑢2 + 2𝑎𝑠 𝑣 2 − 𝑢2 = 2𝑎𝑠 𝑣 2 − 𝑢2 =𝑎 2𝑠 Subtract u2 Divide by 2s Replace a with the expression we worked out 𝑊= 𝑚(𝑣 2 − 2 𝑢2 ) 1 1 𝑚𝑣 2 − 𝑚𝑢2 2 2 1 1 𝑚𝑣 2 − 𝑚𝑢2 2 2 Final KE - Initial KE Multiply all by s Rewrite right side Fs = work done 3B 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 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 0.3kg is moving at a speed of 9ms-1. Calculate its kinetic energy. 𝐾𝐸 = 𝐾𝐸 = 1 𝑚𝑣 2 2 1 (0.3)(9)2 2 Sub in values Calculate 𝐾𝐸 = 12.15𝐽 3B 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 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 1.5kg 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 3ms-1. The work done by the force is 1.8J. Calculate the value of u. We know W, v and m, and need u 𝑊= 1.8 = 1 1 𝑚𝑣 2 − 𝑚𝑢2 2 2 1 (1.5) 3 2 2 1 − (1.5)𝑢2 2 Sub in values Work out parts 1.8 = 6.75 − 0.75𝑢2 Rearrange 2 0.75𝑢 = 4.95 Divide by 0.75 𝑢2 = 6.6 𝑢 = 2.57𝑚𝑠 −1 Square root Use the formula for the change in kinetic energy! 3B 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 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 bus of mass 2000kg starts from rest at some traffic lights. After travelling 400m the bus’s speed is 12ms-1. A constant resistance of 500N acts on the bus. Calculate the driving force, P, which can be assumed to be constant. We know the following pieces of information: u = 0ms-1 v = 12ms-1 s = 400m m = 2000kg We also know the overall force will be the driving force subtract the resistances F = P - 500 500N P 𝑊= 1 1 𝑚𝑣 2 − 𝑚𝑢2 2 2 𝐹𝑠 = 1 1 𝑚𝑣 2 − 𝑚𝑢2 2 2 1 𝑃 − 500 × 400 = (2000) 12 2 Replace W with Fs Sub in values 1 2 − (2000)(0)2 2 400 𝑃 − 500 = 144000 Calculate parts Divide by 400 𝑃 − 500 = 360 Add 500 𝑃 = 860𝑁 3B 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 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 15m. 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 𝑃𝐸 = 𝑚𝑔ℎ 𝑃𝐸 = (30)(𝑔)(−15) Sub in values. The height has fallen by 15m… Calculate 𝑃𝐸 = −4410𝐽 So the loss of potential energy is 4410J 3B 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 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 parcel of mass 3kg is pulled 7m 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. 𝑇𝑎𝑛𝜃 = 3 4 𝑆𝑖𝑛𝜃 = 3 5 𝐶𝑜𝑠𝜃 = 4 5 Draw a diagram 7m θ 7Sinθ 7Cosθ 𝑃𝐸 = 𝑚𝑔ℎ 𝑃𝐸 = (3)(9.8)(7𝑆𝑖𝑛𝜃) 𝑃𝐸 = (3)(9.8) 7 × 3 5 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 𝑃𝐸 = 123.48𝐽 3B 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 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 “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) “The change in total energy of a particle is equal to the work done on the particle.” (This is called the ‘work-energy’ principle) 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = If gravity is the only force acting on a particle: 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 If another force (usually friction) is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 3C 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 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: 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 30° Initial speed = 0 Final speed = 6 Find the increase in Kinetic energy 𝑊= A smooth plane is inclined at 30° to the horizontal. A particle of mass 0.5kg slides down the slope. The particle starts from rest at point A and at point B has a speed of 6ms-1. Find the distance AB. 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = 𝑊= 1 1 𝑚𝑣 2 − 𝑚𝑢2 2 2 1 (0.5) 6 2 2 1 − (0.5)(0)2 2 Sub in values Calculate 𝑊 = 9𝐽 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝐾𝐸 = 9𝐽 3C 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 Work, energy and power 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = Draw a diagram The normal reaction is doing no work as there is no movement perpendicular to x xSin30 the plane The plane is smooth so the 30° particle does not have to do xCos30 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 If gravity is the only force acting on a particle: 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 A smooth plane is inclined at 30° to the horizontal. A particle of mass 0.5kg slides down the slope. The particle starts from rest at point A and at point B has a speed of 6ms-1. Find the distance AB. 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝐾𝐸 = 9𝐽 B 30° Initial speed = 0 Find the decrease in Potential energy (find the change in vertical height first) Call the diagonal distance (the one we need to find) ‘x’ 𝑃𝐸 = 𝑚𝑔ℎ Sub in values 𝑃𝐸 = (0.5)(9.8)(𝑥𝑆𝑖𝑛30) 𝑃𝐸 = 2.45𝑥 Calculate in terms of x 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑃𝐸 = 2.45𝑥 3C 𝑾 = 𝑭𝒔 𝑾= 𝑾 = 𝒎𝒈𝒉 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 Work, energy and power 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = Draw a diagram The normal reaction is doing no work as there is no movement perpendicular to x xSin30 the plane The plane is smooth so the 30° particle does not have to do xCos30 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 If gravity is the only force acting on a particle: 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 A smooth plane is inclined at 30° to the horizontal. A particle of mass 0.5kg slides down the slope. The particle starts from rest at point A and at point B has a speed of 6ms-1. Find the distance AB. 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝐾𝐸 = 9𝐽 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑃𝐸 = 2.45𝑥 B 30° Initial speed = 0 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑃𝐸 = 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝐾𝐸 2.45𝑥 = 9 𝑥 = 3.67𝑚 Sub in the values we calculated Divide by 2.45 This could be calculated using F = ma and the SUVAT equations from M1, however in M2 you will usually be asked specifically to use these principles… 3C 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 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: 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 45° Initial speed = 8 Final speed = 0 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 A particle of mass 2kg is projected with speed 8ms-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. 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = 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 𝑊= 𝑊= 1 1 𝑚𝑣 2 − 𝑚𝑢2 2 2 1 (2) 0 2 2 1 − (2)(8)2 2 Sub in values Calculate 𝑊 = −64𝐽 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 64𝐽 3C 𝑾 = 𝑭𝒔 𝑾= 𝑾 = 𝒎𝒈𝒉 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 Work, energy and power 45° xCos45 If gravity is the only force acting on a particle: 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 xSin45 x You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle 45° Initial speed = 8 Final speed = 0 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 A particle of mass 2kg is projected with speed 8ms-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. 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = 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… 𝑃𝐸 = 𝑚𝑔ℎ Sub in values 𝑃𝐸 = (2)(9.8)(𝑥𝑆𝑖𝑛45) 𝑃𝐸 = 9.8 2𝑥 Calculate in terms of x 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 64𝐽 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 = 9.8 2𝑥 3C 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 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: 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 45° Initial speed = 8 Final speed = 0 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 A particle of mass 2kg is projected with speed 8ms-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. 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = 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 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝐾𝐸 𝑙𝑜𝑠𝑡 − 𝑃𝐸 𝑔𝑎𝑖𝑛𝑒𝑑 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 64 − 9.8 2𝑥 Sub in values to find an expression for the loss of energy 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 64𝐽 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 = 9.8 2𝑥 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 64 − 9.8 2𝑥 3C 𝑾 = 𝑭𝒔 𝑾= 𝑾 = 𝒎𝒈𝒉 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 Work, energy and power 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = 2gCos45 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 7.84Cos45 If gravity is the only force acting on a particle: 45° 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 Initial speed = 8 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 A particle of mass 2kg is projected with speed 8ms-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. 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 64𝐽 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 = 9.8 2𝑥 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 64 − 9.8 2𝑥 Draw a diagram The normal reaction is doing x no work as there is no movement perpendicular to the plane 2gCos45 As the plane is rough, the 45° particle will have to do 2g some work against friction. 2gSin45 You must take this into account. You will need to use the Final speed = 0 second of the formulae to the left The energy lost will all have been used against friction 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 𝐹𝑀𝐴𝑋 = 𝜇𝑅 𝐹𝑀𝐴𝑋 = (0.4)(2𝑔𝐶𝑜𝑠45) 𝐹𝑀𝐴𝑋 = 7.84𝐶𝑜𝑠45 Sub in values Rewrite 3C 𝑾 = 𝑭𝒔 𝑾= 𝑾 = 𝒎𝒈𝒉 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 Work, energy and power 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = 2gCos45 You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle 7.84Cos45 If gravity is the only force acting on a particle: 45° 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 Initial speed = 8 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 Draw a diagram The normal reaction is doing x no work as there is no movement perpendicular to the plane 2gCos45 As the plane is rough, the 45° particle will have to do 2g some work against friction. 2gSin45 You must take this into account. You will need to use the Final speed = 0 second of the formulae to 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 2kg is projected with speed 8ms-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. 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 64𝐽 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 = 9.8 2𝑥 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 64 − 9.8 2𝑥 This is the energy lost 𝑊 = 𝐹𝑠 𝑊 = (7.84𝐶𝑜𝑠45)(𝑥) 𝑊 = 7.84𝑥𝐶𝑜𝑠45 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! 3C 𝑾 = 𝑭𝒔 𝑾= 𝑾 = 𝒎𝒈𝒉 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 Work, energy and power 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = 2gCos45 You can use the principle of the conservation of mechanical energy and the work-energy principle to solve problems involving a moving particle 7.84Cos45 If gravity is the only force acting on a particle: 45° 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 Draw a diagram The normal reaction is doing x no work as there is no movement perpendicular to the plane 2gCos45 As the plane is rough, the 45° particle will have to do 2g some work against friction. 2gSin45 You must take this into account. You will need to use the Final speed = 0 second of the formulae to the left Initial speed = 8 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 A particle of mass 2kg is projected with speed 8ms-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. 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 64𝐽 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 = 9.8 2𝑥 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 64 − 9.8 2𝑥 𝑊 = 7.84𝑥𝐶𝑜𝑠45 Set the two expressions equal to each other and solve for the distance, ‘x’ 64 − 9.8 2𝑥 = 7.84𝑥𝐶𝑜𝑠45 Add 9.8√2x 64 = 7.84𝑥𝐶𝑜𝑠45 + 9.8 2𝑥 64 = 7.84𝐶𝑜𝑠45 + 9.8 2 𝑥 64 7.84𝐶𝑜𝑠45 + 9.8 2 =𝑥 Factorise right side Divide by the bracket Calculate 3.3𝑚 = 𝑥 3C 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 Work, energy and power 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = 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: 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 A skier passes a point A on a ski-run, moving downhill at 6ms-1. After descending 50m vertically, the run starts to ascend. When the skier has ascended 25m to point B her speed is 4ms-1. The skier and skis have a combined mass of 55kg. The total distance travelled from A to B is 1400m. The resistances to motion are constant and have a magnitude of 12N. B 50m 25m 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 loss of kinetic energy (it is a loss as speed has fallen) 𝑊= 𝑊= 1 1 𝑚𝑣 2 − 𝑚𝑢2 2 2 1 (55) 4 2 2 1 − (55)(6)2 2 Sub in values Calculate 𝑊 = −550𝐽 Calculate the work done by the skier 𝐾𝐸 𝐿𝑜𝑠𝑡 = 550𝐽 3C 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 Work, energy and power 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = 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: 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 A skier passes a point A on a ski-run, moving downhill at 6ms-1. After descending 50m vertically, the run starts to ascend. When the skier has ascended 25m to point B her speed is 4ms-1. The skier and skis have a combined mass of 55kg. The total distance travelled from A to B is 1400m. The resistances to motion are constant and have a magnitude of 12N. B 50m 25m 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 gain of potential energy (it is actually a loss as the height has fallen!) 𝑃𝐸 = 𝑚𝑔ℎ Sub in values 𝑃𝐸 = (55)(9.8)(−25) 𝑃𝐸 = −13475𝐽 Calculate Calculate the work done by the skier 𝐾𝐸 𝐿𝑜𝑠𝑡 = 550𝐽 𝑃𝐸 𝐺𝑎𝑖𝑛𝑒𝑑 = −13475𝐽 3C 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 Work, energy and power 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = 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: 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 A skier passes a point A on a ski-run, moving downhill at 6ms-1. After descending 50m vertically, the run starts to ascend. When the skier has ascended 25m to point B her speed is 4ms-1. The skier and skis have a combined mass of 55kg. The total distance travelled from A to B is 1400m. The resistances to motion are constant and have a magnitude of 12N. Calculate the work done by the skier 𝐾𝐸 𝐿𝑜𝑠𝑡 = 550𝐽 𝑃𝐸 𝐺𝑎𝑖𝑛𝑒𝑑 = −13475𝐽 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 14025𝐽 B 50m 25m 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 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝐾𝐸 𝑙𝑜𝑠𝑡 − 𝑃𝐸 𝑔𝑎𝑖𝑛𝑒𝑑 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 550 − (−13475) 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 14025𝐽 Sub in values Calculate It makes sense that these are added together, as we have lost both Kinetic and Potential energies! 3C 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 Work, energy and power 𝟏 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑲𝑬 = 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: 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬 If another force is acting on the particle: 𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅 A skier passes a point A on a ski-run, moving downhill at 6ms-1. After descending 50m vertically, the run starts to ascend. When the skier has ascended 25m to point B her speed is 4ms-1. The skier and skis have a combined mass of 55kg. The total distance travelled from A to B is 1400m. The resistances to motion are constant and have a magnitude of 12N. Calculate the work done by the skier 𝐾𝐸 𝐿𝑜𝑠𝑡 = 550𝐽 𝑃𝐸 𝐺𝑎𝑖𝑛𝑒𝑑 = −13475𝐽 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 14025𝐽 B 50m 25m 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 𝑊 = 𝐹𝑠 𝑊 = (12)(1400) Sub in values – the resistances of 12N act over 1400m Calculate 𝑊 = 16800𝐽 16800J of energy has been used against the resistances. The loss of kinetic and potential energy of 14025J has contributed to this The rest will be work done by the skier 16800 − 14025 = 2775𝐽 3C 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 𝑷 = 𝑭𝒗 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 (kW) where 1kW = 1000W The power developed by an engine is given by the following formula: 𝑷 = 𝑭𝒗 P = power (W) F = the driving force of the engine (N) v = velocity (ms-1) 3D 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 𝑷 = 𝑭𝒗 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 8ms-1 by a force of magnitude 2000N acting parallel to the direction of motion of the truck. Calculate the power developed in kilowatts. 𝑃 = 𝐹𝑣 Sub in values 𝑃 = (2000)(8) Calculate 𝑃 = 16000𝑊 Change to kilowatts 𝑃 = 16𝑘𝑊 3D 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 𝑷 = 𝑭𝒗 Work, energy and power You can calculate the power developed by an engine and solve problems about moving vehicles Draw a diagram and show forces 600N T4000N A car of mass 1250kg is travelling along a horizontal road. The car’s engine is working at 24kW. The resistance to motion is constant and has magnitude 600N. Calculate: v=6 a) 24000 = 𝐹(6) The acceleration of the car when it is travelling at 6ms-1 𝑎 = 2.72𝑚𝑠 −2 b) The maximum speed of the car To calculate the acceleration we can use the formula F = ma. However, we do not know the driving force from the engine yet. We can calculate the driving force from the information given T is often used as the ‘tractive’ force of the engine P = 24000W 𝑃 = 𝐹𝑣 Sub in values Divide by 6 4000𝑁 = 𝐹 𝐹 = 𝑚𝑎 4000 − 600 = (1250)𝑎 Resolve horizontally and sub in values Calculate a 2.72 = 𝑎 At a velocity of 6ms-1, the acceleration is 2.72ms-2 3D 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 𝑷 = 𝑭𝒗 Work, energy and power You can calculate the power developed by an engine and solve problems about moving vehicles A car of mass 1250kg is travelling along a horizontal road. The car’s engine is working at 24kW. The resistance to motion is constant and has magnitude 600N. Calculate: a) The acceleration of the car when it is travelling at 6ms-1 𝑎 = 2.72𝑚𝑠 −2 Draw a diagram and show forces 600N 600N 4000N 𝑃 = 𝐹𝑣 Sub in values 24000 = (600)𝑣 Calculate v 40 = 𝑣 So the maximum speed of the car is 40ms-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 600N! 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 3D 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 𝑷 = 𝑭𝒗 Work, energy and power N T 1600N You can calculate the power developed by an engine and solve problems about moving vehicles R 1100g 7˚ 1100gCos7 A car of mass 1100kg is travelling at a constant speed of 15ms-1 along a straight road which is inclined at 7˚ to the horizontal. The engine is working at a rate of 24kW. As the speed is constant, the driving force must be equal to the forces opposing motion (gravity and anything else) a) Find the driving force first Calculate the magnitude of the nongravitational resistances to motion The rate of working of the engine is now increased to 28kW. Assuming the resistances to motion are unchanged: 𝑃 = 𝐹𝑣 24000 = 𝐹(15) 7˚ 1100gSin7 Sub in values Calculate F 𝐹 = 1600𝑁 b) Calculate the initial acceleration of the car 3D 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 𝑷 = 𝑭𝒗 Work, energy and power N 1600N You can calculate the power developed by an engine and solve problems about moving vehicles A car of mass 1100kg is travelling at a constant speed of 15ms-1 along a straight road which is inclined at 7˚ to the horizontal. The engine is working at a rate of 24kW. a) Calculate the magnitude of the nongravitational resistances to motion 𝑅 = 286.2𝑁 The rate of working of the engine is now increased to 28kW. Assuming the resistances to motion are unchanged: b) Calculate the initial acceleration of the car 268.2N R 7˚ 1100g 7˚ 1100gCos7 1100gSin7 As the speed is constant, the driving force must be equal to the forces opposing motion (gravity and anything else) Now you have the driving force, resolve parallel to the plane 𝐹 = 𝑚𝑎 1600 − 𝑅 − 1100𝑔𝑆𝑖𝑛7 = 0 1600 − 1100𝑔𝑆𝑖𝑛7 = 𝑅 Sub in values. Remember acceleration is 0 Rearrange to find R Calculate 𝑅 = 286.2𝑁 3D 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 𝑷 = 𝑭𝒗 Work, energy and power N 1600N 1867N You can calculate the power developed by an engine and solve problems about moving vehicles A car of mass 1100kg is travelling at a constant speed of 15ms-1 along a straight road which is inclined at 7˚ to the horizontal. The engine is working at a rate of 24kW. a) Calculate the magnitude of the nongravitational resistances to motion 𝑅 = 286.2𝑁 The rate of working of the engine is now increased to 28kW. Assuming the resistances to motion are unchanged: 268.2N 7˚ 1100g 7˚ 1100gCos7 1100gSin7 As the speed is constant, the driving force must be equal to the forces opposing motion (gravity and anything else) For part b), we need to start again by calculating the tractive force of the vehicle 𝑃 = 𝐹𝑣 28000 = 𝐹(15) 𝐹 = 1867𝑁 Sub in values Calculate (remember to use the exact value later on) b) Calculate the initial acceleration of the car 3D 𝑾 = 𝑭𝒔 𝑾 = 𝒎𝒈𝒉 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 𝟐 𝑷𝑬 = 𝒎𝒈𝒉 𝑾= 𝟏 𝟏 𝒎𝒗𝟐 − 𝒎𝒖𝟐 𝟐 𝟐 𝑷 = 𝑭𝒗 Work, energy and power N 1867N You can calculate the power developed by an engine and solve problems about moving vehicles A car of mass 1100kg is travelling at a constant speed of 15ms-1 along a straight road which is inclined at 7˚ to the horizontal. The engine is working at a rate of 24kW. a) Calculate the magnitude of the nongravitational resistances to motion 𝑅 = 286.2𝑁 The rate of working of the engine is now increased to 28kW. Assuming the resistances to motion are unchanged: b) Calculate the initial acceleration of the car 268.2N 7˚ 1100g 7˚ 1100gCos7 1100gSin7 As the speed is constant, the driving force must be equal to the forces opposing motion (gravity and anything else) Now use the F = ma formula again with the updated information… 𝐹 = 𝑚𝑎 1867 − 268.2 − 1100𝑔𝑆𝑖𝑛7 = 1100 𝑎 Sub in values Divide by 1100 0.242 = 𝑎 The initial acceleration will be 0.242ms-2 3D 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