PHYSICS FORM ONE TOPIC 8: WORK ENERGY AND POWER msomimaktaba, November 6, 2018February 13, 2019 WorkThe Concept of WorkExplain the concept of workIf a person pushes a wall and the wall does not move, though the person may sweat and physically become tired, he would not have done any work. But if the person pushes a trolley and the trolley moves it is said work is done.The S.I Unit of WorkState the S.I unit of workWork is the product of force and distance moved in the direction of the force.Thus,Work = Force (f) * distance (d) moved in the direction of the force.SI unit of work is Joules.The Work Done by an Applied ForceDetermine the work done by an applied forceExample 1A sack of maize, which weighs 800N, is lifted to a height of 2m. What is work done against gravity?Data given:Force (f) = 800NHeight distance = 2mWork done = ?Solution: Work done (w.d) = force (f) x distance (d)w.d = 800N x 2m= 1600 JoulesWork done (w.d) = 1600JoulesEnergyThe Concept of EnergyExplain the concept of energyEnergy can be defined as capacity of doing work.Energy has the same SI unit like that of work, and that is Joules (J)S.I Unit of EnergyState S.I unit of energyEnergy has the same SI unit like that of work, and that is Joules (J)Different Forms of EnergyIdentify different forms of energyThere are different forms of energy such as:Chemical energyHeat energyLight energySound energyElectrical energyNuclear energySolar energyDifference between Potential Energy and Kinetic EnergyDistinguish between potential energy and kinetic energyThere are two types of chemical energy, which are:Potential energy:It is the energy possessed by a body mass in its position or state.Kinetic energy: It is the energy possessed by a body due to its motion.Consider when the body is vertically thrown upwards with an initial velocity ‘u’ from the ground.At the ground:The height is zero and initial velocity is at maximum so as to attain maximum height.Therefore K.E = ½ MV2 will be maximumK.Emax = ½ mv2Where K.E = Kinetic energyM = Mass of the object/bodyV=VelvetyP.E = MghWhereP.E = Potential energyM= Mass of the objectH = Height of the objectg = gravitation forceP.E = Mgh will be zero because P.E= M*g*0 (body at the ground where k=0)Neglecting the air resistance, as the body moves upwards its velocity decreases it also experiences gravitational force (g) pulling downwards towards the earth’s centre.The maximum Height AttainedThe final velocity of the body will be zero (V=0)Therefore K.E = ½ mv2 K.E = ½ m(0)2K.E = 0P.E = MgHmaxNote:That the object drops from Hmax that is; it leaves with zero Kinetic Energy. At position A in fig. 8. The conservation of mechanical energy (M.E) is given as:P.E + K.E = Constant(The sum of P.E and K.E is constant throughout the motion of the object if the air resistance is neglected)The Transformation of EnergyExplain the transformation of energyThe notion of energy is that energy is changed from one form into different forms using transducers.Transducer is a device used to transform energy from one form to another.For example:Battery converts chemical energy into electrical energy.A generator converts mechanical energy into electrical energy.A motor converts electrical energy into mechanical energy.The Table Summarising Energy TransformationORIGINAL ENERGYTRANSDUCERENERGY TRANSFORMEDChemical energyBatteryElectrical energyChemical energyMotorChemical energyMechanical energyGeneratorElectrical energySolar energySolar panelElectrical energyChemical energyMotor carMechanical energyElectrical energyMicrophoneSound energyElectrical energyHeaterHeat energyActivity 1To demonstrate pressure of potential energy.Materials and ApparatusA heavy stoneA bucket full of waterA strong inelastic rope; andSmooth pulleyProcedures:Collect the heavy stone, using a strong rope tie it to a bucket of waterPass the rope over smooth pulley fixed to a support.Hold stationary the heavy stone at height “h” above the groundRelease the stoneResults and observations:When the stone released the bucket of water will start to rise.The stone is said to have potentials energy because of its position above the ground.Lifting a body of mass “m” to a height “h” above the ground requires work to be done against gravity. Work = MghExample 2A ball of mass 0.5 kg is kicked vertically upwards and rises to a height of 5m. Find the potential energy by the ball.Data given:Mass of the ball (Mb) = 0.5 kgHeight (h) = 5mGravitation force (g) 10N/kgPotential energy (P.E) = ?Solution:Potential energy (P.E)= mgh= 0.5kg x 10N/kg x 5m= 25 NM1NM= 1JoulesPotential energy (P.E) = 25 Joules.Activity 2Aim: to investigate the law of conservation of energy by a simple pendulum.Materials and Apparatus:A pendulum bob and light inelastic string.ProducersPull the bob of a simple pendulum in position A so that it is at height “h” above position B.Release the bob so that it swings to position C via the lowest position B and back to A.Consider the figure below:ObservationWhen the bob is at position A, it possesses potential energy only due to the height “h” which is equal to “Mgh”.As it swings downwards to position B, the height decreases, and as the result it loses potential energy.The bob has Vmax and hence K.Emax at B. The height at B is zero, thus the P.E is zero.As it swings towards C, the P.E increases and reaches its maximum again in position C, where the Kinetic Energy is zero. At position D, the energy of the bob is party potential and party Kinetic.The Principle of Conservation of EnergyState the principle of conservation of EnergyThe law of conservation of energy state that ” Energy can neither be created nor destroyed but can only be converted from one form to another.”This means the amount of energy is constant all the time.Example 3A stone of mass 2kg is released form a height of 2m above the ground. Find the potential energy of the stone when it is at the height of 0.5m above the ground.Data given;Mass of the stone (Ms) = 2kgHeight released (h) = 2mGravity (g) = 10N/kgPotential energy = (P.E) ?Solution:P.E at height g 2mP.E = Mgh= 21g x 10N/kg x 2mP.E = 40 JoulesP.E at 2m = 40JoulesThan P.E at height of 0.5m= 21g x 10N/kg x 0.5mP.E at 0.5m = 10Joules100s of P.E = 40 Joules – 10Joules= 30 JoulesAccording to conservation of energy the loss of P.E should be equal to the gain in K.E, when the air resistance is neglected.K.E of the stone at 0.5 above the ground = 30 JoulesExample 4A ball of mass 0.21kg is dropped from a height of 20m. on impact with the ground it loses 30J of energy. Calculate the height which it reaches on the rebound.Data given;Mass of ball (Mb) = 0.2kgHeight dropped (h) = 20mLoose in energy (E) 30JHeight which reaches=?Solution;Consider the figure below;At 20m above ground the initial energy of the ball = Mgh= 0.2kg x 10N/kg x 20m= 40 Joules.So after the impact the ball loose 30J and the energy remaining is 40 J-10J= 10JoulesAt the top of rebound the energy of the ball = potential energy (P.E)The height reaches (h) is 5m.Uses of Mechanical EnergyExplain the uses of mechanical energyThe mechanical energy can be used to produce electric power using generators. Some uses of mechanical energy are: It enables our body to do work, it makes work easier and faster, it is used to transport goods and people from one place to another, many transport vehicles uses the knowledge of mechanical energy. Examples of vehicles which uses mechanical energy are aeroplanes and motor cars. PowerThe Concept of PowerExplain the concept of powerPower is the rate of which work is done.It is a measure of the rate at which energy changes.This means that whenever work is done energy changes into a different form.The S.I Unit of PowerState the S.I unit of powerThe SI unit of power is Jules per second J/S or watts, W.1 Joules per second = 1 wattWhen 1 Joules of work is done per second the power produced is a watt. Watt is the unit for measuring electrical power.The Rate of Doing WorkDetermine the rate of doing workSuppose that two cranes each lift objects having masses of 200kg to a height of 12m. Crane A lifts its object in 10sec while crane B requires 15sec to lift its object. Assume they lift the objects at a constant velocity they do the same amount of work.Work done = GPE= Mgh= (200kg) (9.8m/s2)(12m)= 23520JEach did a work that was equivalent to 23520J.What is different for the two cranes is the rate at which they did the work or their generation of power.The power of crane A can be calculated by;PB = 1568 watts.Example 5How much power is required to accelerate a 1000kg car from rest to 26.7m/s in 8sec?Solution:The work done on the car increases its Kinetic energy.Work done = AKE½ MV2 – ½ MV2The power required is given by:Example 6Car engine is rated in horsepower (hp) where 1hp = 746watts. What is the required power measured in horsepower?Since work causes a change in energy. DE power can be considered as the rate of change of energy.P = DE/t ALL NOTES FOR ALL SUBJECTS QUICK LINKS:AGRICULTURE O LEVEL PURE MATHEMATICS A LEVELBAM NOTES A LEVELBASIC MATH O LEVELBIOLOGY O/A LEVELBOOK KEEPING O LEVELCHEMISTRY O/A LEVELCIVICS O LEVELCOMPUTER(ICT) O/A LEVELECONOMICS A LEVELENGLISH O/A LEVELCOMMERCE O/A LEVELACCOUNTING A LEVELGENERAL STUDIES NOTESGEOGRAPGY O/A LEVELHISTORY O/A LEVELKISWAHILI O/A LEVELPHYSICS O/A LEVELMOCK EXAMINATION PAPERSNECTA PAST PAPERS Form 1 Physics Notes Msomi Maktaba All Notes Physics Study Notes FORM 1PHYSICSPost navigationPrevious postNext postRelated Posts Form Six Past Papers Arabic Form Six Past Papers – NECTA A Level October 16, 2019October 16, 2019Click The Links Below to download Form 6 Past Papers Arabic Year Questions/Answers 2018 Paper 1 2017 Paper 2 2016 Paper 1, Paper 2 2013 Paper 1 A LEVEL NECTA PAST PAPERS – Form Six (ACSEE) Powered BY maktaba.tetea.org ALL NOTES FOR ALL SUBJECTS QUICK LINKS: AGRICULTURE O… Read More Form One Selection 2024 Rukwa February 2, 2024Form One Selection 2024 Rukwa – List of Selected students to join form one studies for 2023/2024 intake Rukwa This is to inform the general public that selection process for admission into secondary school – Form one studies for Government Secondary Schools for 2024 Academic Session is still going on…. 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