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Lecture 5 Work Energy Work, Energy Work and energy are fundamental physical quantities in science. Work is done when a force moves an object through a distance. Energy is the ability to do work The unit of both work and energy is the Joule. James Joule 1818—1889 English Physicist 4186 J = 1kcal Work F distance (s) The work done by a force F moving an object by a distance s in the direction of the force is defined as: W = Fs units of Newton.metre (Nm) = Joule If the force points in a direction different to that of the displacement, the work done is given by the component of the force parallel to the direction of displacement multiplied by the displacement. Work If the force points in a direction different than the displacement F q FCosθ displacement (s) W = (FCosθ)s Only the component of force parallel to the direction of displacement does work on the object. Both work and energy are scalars. Energy is always positive. Work is positive if both force (or component) and displacement point in the same direction Work is negative if they point in opposite directions. Work Work done only when energy is transferred into or out of a system Examples: Horizontal motion: Vertical motion: object carried F mg The work done by the d weight is: W=Fd =(mg)d d mg F mg The work done by the force F is: W = (FCosθ)d W=mg*0*d = 0 J When force and motion are perpendicular to each other the work done is zero Work is in the direction of force In general, only the projection of the force in the direction of the displacement does work: W = FHd Cos30o = FH ÷ F FH = FCos30o FH =10*(0.866)N FH = 8.66N W = FCos30o *d = FH*d = 8.66*d joules Work Example How much work is done on an object of mass 10kg (a) As it’s lifted straight up through a distance of 1m (b) as it’s lowered straight down through a distance of 1m. W = (FCosθ)s W=(FCos0)s=Fs (a) F=mg finish W=mgs=10kg(9.8ms-2)1m mg W=98N.m = 98J s start (lifting) Work is positive since the force is in same direction as movement Example contd (b) as it’s lowered straight down through a distance of 1m. W = (FCosθ)s start s W=FCos(180o)s = -Fs F=mg finish mg Force & s are in opposite directions W = -mgs = -10kg(9.8ms-2)1m W = -98N.m = -98J (lowering) Work is negative since the applied force is in opposite direction to movement Work is transfer of energy V = 0km/h V = 100km/h W The engine of the car does work: It changes the ‘chemical’ energy of the fuel into motion Work is transfer of energy Energy Sun is the ultimate source of energy on Earth Heat and light Almost all life on Earth is dependent on the Sun We get energy from food Food grows by capturing sun’s energy by a process called photosynthesis Solar energy →chemical energy Whatever we eat, energy within it comes from the sun Fossil fuels: turf, coal, oil, gas, all forms of energy Energy needed to form these came from the sun Fossil fuels are made from plants and animals that lived and died in swamps millions of years ago. Energy (J) Energy is the ability to do work. Work done by a system takes energy out of the system Work done on a system puts energy into it. Energy can exist in many forms: -Kinetic energy: the energy linked to movement -‘stored’ energies: elastic energy, chemical energy, electrical energy, gravitational potential energy,… -‘Thermal energy: proportional to temperature -Radiation energy: electromagnetic waves (light, etc…) Work and Energy Consider work done on a system, producing motion. If all the work done goes into causing motion then the energy of motion is equal to the work done. From Kinematics v v 2as 2 2 0 v 2 v02 a 2s If we multiply a by the mass m of the object mv 2 mv02 F ma 2s mv 2 mv02 Fs 2 2 mv 2 Fs 2 If initial speed is zero Work and Energy mv 2 Fs 2 Fs is just the work done on the object by the applied force F. The quantity 1 2 mv 2 is known as the Kinetic energy (KE) of a object with mass m traveling at speed v. 1 2 KE mv 2 Kinetic Energy Kinetic energy is the energy associated with an object in motion: 1 2 KE mv 2 m: mass of the body v: speed of the body Example: V = 0km/h V = 100km/h Mass of the car: 1ton =1000kg Calculate final KE and work done by engine Initial kinetic energy: KEinit = ½*1000*02 J = zeroJ Final kinetic energy: KEfinal = ½*1000*(100/3.6)2 J = 386kJ The work done by the engine is: W = 386kJ Kinetic energy Example A father (mf=80kg) is pushing a buggy (mb=20kg). He exerts a force of 10N directed 30° below the horizontal. Assume friction is negligible. What is their speed after 2m, starting from rest? W = FHd = F*cos(30°)*d = 8.66N*2m = 17.32J Einit = 0J Efinal = ½(mf+mb)v2 Efinal = W = 17.32J 1 17.32Joules= (m f mb )v 2 2 v 2 17.32 2 17.32 0.59m / s m f mb 80 20 v 0.59ms 1 Potential Energy We have seen that Kinetic energy is the energy related to an object’s motion. There is another type of energy which is related to a object’s shape or position, this is known as Potential energy (PE). Examples: Shape: Coiled Watch spring, Position Water at the top of a hydroelectric dam We will consider potential energy mainly related to position. Potential Energy Example: A man lifts a box of mass m from the ground to a height h. What is the work done on the box? h Work done = energy given to box W = force x distance = Fh W = mgh Initial and final kinetic energy is zero (v=0) Energy of box is that associated with its position Gravitational potential energy PE = mgh Potential Energy Example A man lifts a box of mass 20kg from the ground to a height of 1.5m. What is the work done on the box? Initial potential energy: PEinit = 20*9.8*0 J = 0 J Final potential energy: PEfinal 20*9.8*1.5J = 294J The work done by the man is: W = 294J PE = m g h J kg m.s-2 m: mass of the body h: altitude of the object m Linking KE and PE Total energy, kinetic, potential, chemical etc in a closed system is constant or conserved. This is an example of the principle of conservation of energy. System is closed if no energy enters or leaves it by any method Energy can be transferred only between components within the system principle of conservation of energy KE +PE+OE =constant OE is the sum of all other forms of energy Energy can be transferred within the system Initial total energy equals the final total energy KEi +PEi+OEi = KEf +PEf+OEf Linking KE and PE Initial energy = Final energy (1/2)mvi2 + mghi +OEi = (1/2)mvf2 + mghf +OEf initial kinetic energy initial initial all final kinetic potential other energy energy energy final potential energy Final all other energy Just consider KE and PE then (1/2)mvi2 + mghi = (1/2)mvf2 + mghf In other words the sum of Kinetic and Potential energy is the same before and after the force acted on the object. Linking KE and PE An object drops vertically, starting with zero velocity. If air resistance (friction) is negligible, what is its speed after 10m? mg h Initial kinetic energy = ½mvi2 = 0 Initial potential energy =mgh Potential energy decreases: Kinetic energy increases: energy transfered from potential to kinetic: Using conservation of energy KE = PE mgh = ½mv2 v 2 gh v v 2as 2 2 0 v 2as v = 14m.s-1 Linking KE and PE A sledge with zero initial velocity slides down a bumpy slope. What is its final speed, after loosing 5m of altitude? Assume friction is negligible h KEi +PEi+OEi = KEf +PEf+OEf Since there is no friction OEi = OEf 0 + PEi = KEf + 0 KE= ½mv2 PE= mgh v 2gh mgh = ½mv2 v =√2*9.8*5 ms-1 v = 9.9m.s-1 This approach works whatever the complexity of the trajectory. Conservation of energy If the final velocity of the sledge (10kg) is 8 ms-1, instead of 9.9 ms-1? How much energy is converted to thermal energy by friction on the way down? KEi +PEi+OEi = KEf +PEf+OEf 0 + mghi +0 =1/2mvf2 + 0 + thermal energy Thermal energy = mgh -1/2mvf2 = 10*9.8*5 - ½*10*82 = 170J Friction transfers mechanical energy into thermal energy Thermal energy is another form of energy: associated with the random kinetic energy of atoms and molecules in an object. Roller coaster? Exercise: A roller coaster, starting with 1m/s, 20m above ground. What is its final speed if friction is neglected? Start 20m End KEi +PEi+OEi = KEf +PEf+OEf (1/2)mvi2 + mghi = (1/2)mvf2 + mghf (1/2)mvi2 + mghi = (1/2)mvf2 + 0 2 2gh v vinit v = 19.8ms-1