1. JR/2008 Work & Energy If an object moves due to the action of an applied force the force is said to have done WORK on the object. Work is the product of force applied and the distance moved in the direction of the force. WORK DONE = FORCE x DISTANCE The unit of work can be found from Force = Newtons Distance = metres Work is therefore measured in Newtons x metres or Newton. metres However:- Work is usually defined in units of Joules (J)

2. JR/2008 Work & Energy The Joule is defined as:- The amount of work done when a force of 1 Newton acts for a distance of 1 metre. 1 Joule = 1 Newton metre 1 J = 1 N.m If a graph is plotted of Force ( vertical y axis) against : distance ( horizontal x axis) the force/distance graph is given. This graph is known as a WORK DIAGRAM The area beneath the graph equals the work done

3. JR/2008 Work & Energy The work diagram a.Constant force b.Changing force Force F Work done = area of graph WD = F.s (Nm or J) Work done = area of graph WD = ½ F.s (Nm or J) Distance (s) Force F Distance (s)

4. JR/2008 Work & Ener gy Ex1A spring is stretched 25mm by a force of 50N. Draw a diagram of the work done over the next 50mm of stretch and from this find the work done in that 50mm and the average force applied Solution:- Spring stiffness = 50N/25mm = 2N/mm = 2kN/m The next 50mm needs 50x 2N/mm = 100 N Work done = area of graph WD = ½ (50 + 150) x 50 x10 -3 = 5 Nm or 5 J Average force = (50 + 150)/2 = 100N Force N Distance (mm) 7525 50 150

5. JR/2008 Work & Energy Examples: 1.Calculate the work done in lifting a mass of 0.5 tonne through a vertical distance of 10m Soln. Work = force x distance In this case the work is done against gravity – thus force = weight (mg). When:- mass = 0.5 tonne = 500 kg, g = 9.81, distance = height 10m Work = mass x g x height = 500 kg x 9.81 x 10 = 49050 J = 49.05 kJ

6. JR/2008 Work & Energy Examples: 2.A sledge is pulled along horizontal ground at a constant speed by a rope which makes an angle of 25° with the horizontal. If the tension in the rope remains constant at 100N calculate the energy supplied in moving the sledge 1 km. Soln. Work = force x distance In this case the work is done in the horizontal direction. To find the force in this direction we must resolve the tension (force) in the pulling rope. Horizontal pull = T cos 25° = 100 cos25° =90.63 N Work = force x distance = 90.63 x 1000m Work = 90.63 kJ (in the horizontal direction) 25° Tension T Horizontal pull = T cos 25 °

7. JR/2008 Examples: 3.A force acts on a body of mass 15kg, which is initially at rest, and imparts a constant acceleration of 2 ms -2 for 10 seconds. Calculate the energy transfer (work done ) to the body. Soln. Work done = Energy used = force x distance In this case the equations of motion must be used to find distance and force. u = 0 ms -1 v = ? a = 2 ms -2 s = ? t = 10 seconds Work & Energy From s = ut + ½ at 2 distance s = ½ x 2 x 10 2 = 100m From N2 F = ma, Force = 15 kg x 2 ms -2 = 30N Work done = Energy transferred = 30N x 100m = 3000 J Energy transfer = 3 kJ

8. JR/2008 Work & Energy Energy is defined as the ability to do work WORK DONE = ENERGY USED The unit of ENERGY is as work:- Newtons x metres or Newton. Metres or Joules (J) Energy is given up when work is done

9. JR/2008 Work & Energy Typical forms of energy are: Mechanical Electrical Heat Chemical Nuclear Light Sound Energy may be converted from one form into another In doing so its value remains the same

10. JR/2008 Work & Energy Conservation of Energy:- the principle states that:- ENERGY CANNOT BE CREATED OR DESTROYED Typical conversions include:- Mechanical into electrical by a generator Electrical into mechanical by a motor Heat into mechanical by a steam turbine Chemical to heat by combustion Nuclear to electrical via a steam turbine & generator Sound to electrical by a microphone Many other conversions are regularly seen, try to discover an area where energy is simply ‘lost’ or created from nothing.

11. JR/2008 Work & Energy The conversion of energy is never simple – and the conversion is never 100 % efficient, losses will always occur. A loss of Energy does not mean a reduction in the overall quantity but a conversion into a form which is not required. A simple example is the petrol engined motor car. The fuel (chemical energy) is intended to provide motion. In fact only 25% of the energy within the fuel provides motion, the remaining 75% is converted to heat, vibration, sound etc. The car is said to be 25% efficient. Great efforts are made in engineering to try and improve the efficiency of energy conversion.

12. JR/2008 Work & Energy The conversion process can be simplified with a block diagram:- Input Energy = Output Energy + ‘Lost’ Energy EFFICIENCY =OUTPUTx100% INPUT LOSSES PROCESSOUTPUTINPUT

13. JR/2008 Work & Energy Using the equation:- EFFICIENCY = OUTPUTx100% INPUT It can be seen that output will always be < input thus efficiency will always be < 100% The symbol used for efficiency is η (eta) and is given as a percentage Ex1.An electric motor takes 5000 J from the mains supply. If the output shaft of the motor provides 3600J find the efficiency of the motor. Soln.Efficiency = Output/Input x 100% η = (3600 J ÷ 5000 J) x 100% = 72% the 28% lost energy is probably converted to heat

14. JR/2008 Work & Energy Ex2.A petrol engine has an efficiency of 32% and delivers 12.5 kJ/second at the crankshaft. What is the energy input to the engine. Soln. Efficiency = output/input 32/100=12.5 kJ/s ÷ Input Input=12.5 ÷ 0.32 kJ/s = 39.0625 kJ/s Engine η = 32% Input 12.5 kJ/s

15. JR/2008 Work & Energy Ex3.A motor has an efficiency of 75% and drives a gearbox with an efficiency of 87%. If the output from the gearbox must be 20.76 kJ/s calculate the motor input in kJ/s Soln. Method 1 : - interim stages Find output from motor ‘A’ kJ/s; (this must be the required input to the gearbox). A = 20.76/ 0.87 = 23.862 kJ/s which must be the motor output Motor input = 23.862 kJ/s / 0.75 = 31.82 kJ/s motor η = 75% Input20.76 kJ/s gearbox η = 87% ‘A’ kJ/s

16. JR/2008 Work & Energy Soln. Method 2 : - single stage Motor input = 20.76 kJ/s / (0.87 x 0.75) = 31.82 kJ/s Note: 87 % = 87/100 = 0.87 motor η = 75% Input20.76 kJ/s gearbox η = 87%

17. JR/2008 Work & Energy POTENTIAL ENERGY PE This is the energy possessed by a body due to its position in space The amount of energy held will depend upon the height of the body above a specified datum. Work done = Force x distance WD = Energy = Weight x height Energy = mg x h Potential Energy (PE) = mgh When m = mass kg g = 9.81 ms -2 h = height m CoG datum Height h Above datum

18. JR/2008 Work & Energy KINETIC ENERGY KE This is the energy possessed by a body due to its motion in space The amount of energy held will depend upon the mass and the speed of the body. For this to occur then Newton’s 1 st law suggests an opposing force must be applied. The force needed to cause the deceleration = ma The rate of deceleration = v/t thus Force needed = m v/t The distance moved in this period = ½ vt As Work = Force x distance= m v/t x vt/2 = ½ mv 2 KINETIC ENERGY = ½ mv 2 Consider mass m kg, moving at velocity v ms -1 which is brought to rest in a time of t seconds

19. JR/2008 Work & Energy Examples 1.A crane requires 29.5 kJ of energy to raise a mass of 500 kg to a certain height. Determine the height of the load after the crane has stopped lifting. Soln. The crane will do work lifting the mass against gravity (weight) and the load will gain potential energy mgh From PE = mgh the height gain h = PE/mg When PE = 29.5 kJ = 29500 J m = 500 kg g = 9.81 ms -2 Height h = 29500/ (500 x 9.81) = 6.01 metres

20. JR/2008 Work & Energy Examples 2.A motor car of mass 900 kg is travelling at 50 km/hr Determine: a. The kinetic energy of the vehicle b. The speed of the car after it has made a brake application in which 45 kJ of energy is lost Soln. a.KE = ½ mv 2 when:m = 900 kg v = 50 x (1000/3600) = 13.889 ms -1 KE = ½ x 900 x 13.889 2 = 86806.9 J= 86.81 kJ b.If the brakes take out 45 kJ of energy the car is left with:- 86.81 kJ - 45 kJ = 41.81 kJ this is in the form of KE ( ½ mv 2 ) Transpose for v = = = 9.64 ms -1 (34.7 km/hr)

21. JR/2008 Work & Energy Examples 3.A body of mass 800 kg is hauled up an incline plane through a vertical distance of 20m. If the work done against friction is 12 kJ calculate the total work done and the potential energy of the body at its elevated position. Soln. (tip - work out the second part first) Work done lifting a height of 20m is equal to the potential energy gain PE = mghwhen:- m = 800 kg g = 9.81 ms -2 h = 20m PE = 800 x 9.81 x 20 = 156960 J = 156.96 kJ BUT 12 kJ of energy was used to overcome the friction on the incline Thus total energy used to lift = 156.96 + 12 = 168.96 kJ

22. JR/2008 Work & Energy Examples 4.A railway train of mass 300 tonnes is travelling at 80 km/hr when its brakes are applied to reduce its reduce its speed to 35 km/hr Determine the amount of energy used in the brake application. Soln. Initial KE = ½ mv 2 when:m = 300 tonnes = 300 x 10 3 kg v = 80 x (1000/3600) = 22.222 ms -1 KE = ½ x 300x10 3 x 22.222 2 = 74072592.6 J = 74.07 MJ Final KE = ½ mv 2 when:v = 35 x (1000/3600) = 9.722 ms -1 KE = ½ x 300x10 3 x 9.722 2 = 14177592.6 J = 14.18 MJ The energy taken out by the brakes is the ‘loss’ in Kinetic Energy Change in energy (∆ KE) = Initial – Final = 74.07 – 14.18 = 59.89 MJ