Understanding Kinetic Energy and Work-Energy Theorem in Physics

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University of Saskatchewan**We aren't endorsed by this school

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PHYSICS 117

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Physics

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Dec 11, 2024

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Physics 115 - 04Physics 115 (04)Lecture 5Tue 8 Oct 2024

All students are welcome to attend any or all sessionsPHYS 115 Structured Study SessionsWednesday 4:00 - 5:20 pmThorv 205AJehanTuesday 5:00 - 6:20 pmGeology 255 *new locationGraysonMonday 1:30-2:50 pmRoom G3, Murray LibraryAndrewMidterm Review Session Sat. Oct. 12th, 20241 pm, onlinephys_sss@usask.caLink to session will be on the Canvas site for Phys 115 SSScanvas.usask.ca/courses/105817Thursday 5:00 - 6:20 pmRoom G3, Murray LibraryJadaWe will try to record the session, but can’t guarantee there will be a recording.

Physics 115 - 04Today’s Scheduled Topics§5.2 Kinetic Energy and the Work-energy Theorem§5.4 Gravitational Potential Energy§5.5 Gravity and Nonconservative Forces§5.9 Work Done by Varity Forces§5.6 Spring Potential Energy§5.7 Systems and Energy Conservation§5.8 Power§6.1 Momentum and Impulse

Physics 115 - 04Net WorkThe net work done on an object is the sum of the works done by each force on the object.i.e. WNet= W1+ W2 + W3+ …This is equal to the work done by the net force.Proof: Work done by net force is:Define x-axis to be in displacement direction:=FFNetcosNetxW=FFxF()xFWx=Net)(xFx== sum of the works done by each force.=W+x

Physics 115 - 04Work-Energy TheoremConsider the work done by the Net Force on an object. We know: For simplicity let the motion be just along the x-axis.In 1-dimension:And let the net force be constant.So:=aFm=xxmaFaaFxx===constantconstant ()netxWFxma x==xFxmaxThen: i.e. and are only along -axisxaF

Physics 115 - 04Work-Energy TheoremFrom: (since a= constant)Then:This result is true even if:(Can be proved using calculus)()202212022vvxaxavv−=+=()202122120221netmvmvvvmW−=−=constantxFxFxmaxv0v()netxWFxma x==

Physics 115 - 04Work-Energy TheoremThe result is true even if the motion is not in 1-dimension:i.e.This is a very fundamental and general result!221221netifmvmvW−=vi= initial speedvf= final speedIf net work is done on an object, then there is a change in the quantity 12𝑚𝑣2equal to the net work.

Physics 115 - 04Work-Energy TheoremDefine: Kinetic Energy = Energy of motionThen:Work –Energy Theoremi.e. Net Work = Change in Kinetic EnergyNotes:SI unit for Kinetic Energy = Joule (1 J = 1 N.m = 1 kg.m2/s2)KEdepends only on the speed of the mass m.(Does not depend on the direction of the velocity.)221mvKE=KEKEKEWif=−=net

Physics 115 - 04Kinetic Energies of Some ObjectsObjectMass (kg)Speed(m/s)Kinetic Energy (J)Earth orbiting the Sun9.98 10242.98 1052.66 1023Car at 110 km/h2 000309 105Running Athlete70103 500Stone dropped 10 m1.01498Raindrop at terminal speed3.5 10−59.01.4 10−3Oxygen molecule in air5.3 10−265006.6 10−21

Lecture Example 25A box, of mass 4.0 kg, slides down a frictionless incline, which is at an angle of 30to the horizontal. It slides a distance of 2.5 m. (a) What is the net work done on the box as it slides? (b) If the box starts from rest, what is its final speed? m

A box, of mass 4.0 kg, slides down a frictionless incline, which is at an angle of 30to the horizontal. It slides a distance of 2.5 m. (a) What is the net work done on the box as it slides? (b) If the box starts from rest, what is its final speed?

The work required to accelerate an object on a frictionless surface from a speed 𝑣to a speed 2𝑣is...(A) four times the work required to accelerate the object from 2𝑣to 3𝑣.(B) not known without knowledge of the acceleration.(C) twice the work required to accelerate the object from 𝑣= 0 to 𝑣.(D) equal to the work required to accelerate from 𝑣= 0 to 𝑣.(E) three times the work required to accelerate the object from 𝑣= 0 to 𝑣.222net111(2 )3()222fiWKEKEKEmvmvmv=−= =−=

Physics 115 - 04Potential Energye.g. A raised massAs the mass falls the gravitational force does work on it.We might say: The Gravitational Force has the potential to do work on the raised mass.v= 0vymg

Physics 115 - 04Gravitational Potential EnergyWork done by the force of gravityGroundy = 0yi)(gravfiyymgrmgW−==yfmgyiyfmgcosgravrmgW=Path of boxrr)(fiyymg−=Work depends only on the force and displacement Therefore work done by gravity it is independentof the actual path taken.rrcos

Physics 115 - 04Gravitational Potential EnergyIn all cases the work done by the force of gravity does not depend on the pathIt depends only on the change in the box-earth distance.We can define:Gravitational Potential EnergySo:constant)i.e.Earth (near ==gmgyPE()gififWmg yymgymgy=−=−PE = 𝑃𝐸𝑓− 𝑃𝐸𝑖= Change in Gravitational Potential EnergyifPEPE=−PE= −()fiPEPE= −−

Physics 115 - 04Gravitational Potential EnergyThe gravitational potential energy keeps track of the work done by the gravitational force.As a mass moves down:Gravity does positive workThere is a decrease in gravitational PEAs a mass moves up:Gravity does negative workThere is an increase in gravitational PEgWPE= −gFmg=rgFmg=r

Physics 115 - 04Conservative ForcesThe gravitational force is an example of a class of forces known as Conservative Forces.For all conservative forces:•The work done on a particle depends onlyon the initial and final positions –not on the path taken.mgyPEPEPEPEmgymgyWiffig=−=−−=−=)(i.e. for Gravity:+yFg= mg

Physics 115 - 04Since the Work done by conservative forces depends only on the initial and final positions,it is always possible to define a Potential Energy, PESuch that:e.g. When a book is raised, it is always possible to recover the negative work done on it by gravity by allowing the book to fall, then gravity does positive work on the book which becomes kinetic energy.Conservative ForcesccWPE−=Conservative Forces only.

Physics 115 - 04NonconservativeForceThe work done on a particle DOES depend on the path.e.g. Friction (e.g. book sliding on table top)kfFriction force is always in the opposite direction to the direction of motion.Therefore, a longer path means a larger magnitude work is done by friction.rkfrkf1rkf2rIn that case the work done does not depend only on initial and final positions.Therefore, it is impossible to define a potential energy.

Physics 115 - 04Work done by FrictionA friction force is always in the opposite direction to the motion.e.g.Physics book 0=vvPhysics book kfr)180cos(Friction=rfWkrfWk−=FrictionTherefore, the work done by friction is always negative.Friction transfers energy from the system into other forms e.g. sound or heat. This energy is not recoverable by the system. (i.e. it is a nonconservative force)

Lecture Example 26The driver of a car moving at a speed of 50.0 km/h on a level road, slams on the brakes and the car skids to rest. If the coefficient of kinetic friction between the tires and the road is 0.500, what is the stopping distance?  km/hd

The driver of a car moving at a speed of 50.0 km/h on a level road, slams on the brakes and the car skids to rest. If the coefficient of kinetic friction between the tires and the road is 0.500, what is the stopping distance?

Physics 115 - 04Mechanical EnergyWe can write the net work done:nccnetWWW+=Wnc= Work done by nonconservative forcesKEW=netFrom the Work-Energy theorem:For a conservative force:PEW−=cTherefore:KEWW=+nccKEWPE=+−ncifKEKEKE−=ifPEPEPE−=ififPEPEKEKEW−+−=ncSo:PEKEW+=nc)()(nciiffPEKEPEKEW+−+=Or:Wc= Work done by conservative forces

Physics 115 - 04Mechanical EnergyWe define: Total Mechanical Energy to be kinetic energy plus potential energy.i.e.Then:If nonconservative forces do no work,i.e. Wnc= 0,then we haveConservation of Mechanical EnergyPEKEE+=EEEWif=−=ncfiEE=)()(nciiffPEKEPEKEW+−+=

Physics 115 - 04Gravitational Potential EnergyGravitational Potential Energy:yis the height above some arbitrary position where we choose PE= 0.Any choice of y= 0is possible since only changes in potential energy are important in the energy equations.e.g.is the same regardless of where we choose y= 0. mgyPE=)(ifififyymgmgymgyPEPEPE−=−=−=

Lecture Example 27A box, of mass 4.0 kg, slides down a frictionless incline, which is at an angle of 30to the horizontal. It slides a distance of 2.5 m. If the box starts from rest, what is its final speed? m

A box, of mass 4.0 kg, slides down a frictionless incline, which is at an angle of 30to the horizontal. It slides a distance of 2.5 m. If the box starts from rest, what is its final speed?

Physics 115 - 04Work Done by a Varying ForceSimplify: one dimensionxmFxxxfFxxixfxiTotal work going from xito xfis++xFxFW21This approximation gets better as we allow the size of the x segments to get smaller.In the limit:W= area under the Fxverses xgraph.

Physics 115 - 04Spring ForceThe force exerted by a spring is, to a good approximation, proportional to the extension or compression of the spring.Relaxed position orEquilibrium position(not stretched or compressed)+xx = 0FsFskxFs−=An ideal spring obeysHooke’s Lawk= Spring Constant

Physics 115 - 04Spring ForceMinus sign indicates that the force is in the opposite direction to the displacement from the equilibrium position.This is an example of a “Restoring Force”–one that always acts toward the equilibrium position.Equilibrium position+xx = 0FskxFs−=Hooke’s Lawk= Spring ConstantHooke’s Law Simulation

Physics 115 - 04Work done by a Spring+xx = 0xixfFs+xxixf−kxi−kxfWhat is the work done by the spring force on the mass as it moves from xito xf?= Ws= Area under force vs displacement graph.=2212212121))(())((fiiiffskxkxkxxkxxW−=−−−=FsFsArea(0 →xf)−Area(0 →xi)

Physics 115 - 04Elastic Potential EnergyThe spring force is another example (like gravity) where the work done by the spring depends only on its initial and final position.i.e. It is a conservative force.It is therefore possible to define a potential energy stored in the spring.This is called Spring Potential Energy (or Elastic PE)We must have:+xx = 0xixfFsFs221221fiskxkxW−=ssWPE−=We found:Depends only on xiand xf

Physics 115 - 04Elastic Potential EnergyTherefore, since:221221ifsskxkxWPE−=−=221221fiskxkxW−=So we can define, for an ideal spring (or Hooke’s Law spring):221kxPEs=Where: PE= 0when x= 0(relaxed spring)In many cases there will be changes in both spring potential energy and gravitational potential energy.i.e. sgPEPEPE+=Hooke’s Law Energy SimulationMass on vertical spring simulation.

Lecture Example 28When a mass of 0.200 kg is hung from the end of a vertically suspended spring, the spring is stretched by 10.0 cm.(a) What is the spring constant of the spring?(b) How much work is done by the spring force on the mass as it stretches through this distance?10.0 cm

When a mass of 0.200 kg is hung from the end of a vertically suspended spring, the spring is stretched by 10.0 cm.(a) What is the spring constant of the spring?(b) How much work is done by the spring force on the mass as it stretches through this distance?

Lecture Example 29A mass of 1.00 kg is dropped from a height above the floor of 80.0 cm onto a spring, of unstretched length of 20.0 cm, which is resting on the floor. The mass comes to rest, momentarily, at a height of 10.0 cm from the floor. If we can ignore friction, find the spring constant of the spring. 20.0 cm10.0 cm80.0 cm

A mass of 1.00 kg is dropped from a height above the floor of 80.0 cm onto a spring, of unstretched length of 20.0 cm, which is resting on the floor. The mass comes to rest, momentarily, at a height of 10.0 cm from the floor. If we can ignore friction, find the spring constant of the spring.

Lecture Example 2920.0 cm10.0 cmA small block of mass 5.80 g is placed against a compressed spring with a spring constant of 6.00 N/m as shown in the figure below. The spring is compressed a distance x = 9.80 cm. When the spring is released, the block is pushed to the right, and it travels a distance d = 72.2 cm before coming to rest.(a) Calculate the maximum speed of the block just after it leaves the spring (before the friction slows it down). (b) Calculate the work done by the friction as the block slides along the surface. (c) Determine the coefficient of kinetic friction between the block and the surface. If you did not obtain an answer for (b), use a value…, xd

An object on a frictionless surface is pushed against a horizontal ideal spring, so that the spring is compressed a distance d. The object is released, and it has a kinetic energy of KE1 when it loses contact with the spring. The object is then pushed against the spring so that it is now compressed a distance of 3d. Which one of the following expressions is correct for the kinetic energy, KE2, of the object when it loses contact with the spring? (A) KE2 = 2KE1 (B) KE2 = 3KE1 (C) KE2 = KE1 (D) KE2 = 9KE1 (E) KE2 = 1.41KE1Lecture Example

Physics 115 - 04Energy –SummaryDefine Work:Define Kinetic Energy:Work-Energy Theorem:Conservative Forces (work done depends only on initial and final positions): Can Define Potential Energy PE:e.g. Gravitational PE:e.g. Spring PE:Non-Conservative Forces: e.g. Friction:Define Mechanical Energy: cosFdW=𝑑Ԧ𝐅221mvKE=KEW=netccWPE−=Earth)(near mgyPE=rfWk−=FrictionPEKEE+=EEEWif=−=ncfiEE=0Ifnc=Wmequilibriuat 0221==xkxPEsConservation of Mechanical Energy

Lecture Example 30A child on a sled starts from rest at the top of a hill that is inclined at = 10.5to the horizontal. From the top to the bottom of the hill is d= 200. m and the coefficient of kinetic friction between the sled and the snow is 0.0750. At the bottom of the hill, the snow is level and the coefficient of kinetic friction is the same. What distance, x, will the sled slide on the horizontal part of the snow before coming to rest?xd

A child on a sled starts from rest at the top of a hill that is inclined at = 10.5to the horizontal. From the top to the bottom of the hill is d= 200. m and the coefficient of kinetic friction between the sled and the snow is 0.0750. At the bottom of the hill, the snow is level and the coefficient of kinetic friction is the same. What distance, x, will the sled slide on the horizontal part of the snow before coming to rest?

Physics 115 - 04PowerPower is the rate of energy transfer.Energy transfer can be when one object does work on another:e.g. pushing bookIf Work Wis done in a time interval tAverage Power:Instantaneous Power:Now: So:Average Power: Instantaneous Power:tWP=tWPt=→0limconstantIfcosFxFW=costxFP=cosFvP=is the angle between displacement ∆𝐱and Ԧ𝐅cosvFP=is the angle between velocity 𝐯and Ԧ𝐅Physics book

Physics 115 - 04PowerIn general, for any kind of energy transfer:where energy Eis transferred in time interval t.SI Unit:Watt (W) 1 W = 1 J/sNamed after James Wattinventor of the steam engine.Another common unit: horsepower (hp)1 hp = 746 WEPt=Image: CC0 Creative Commons

Lecture Example 31A car delivers 100. hp to the wheels. The mass of the car and driver is 1800. kg. If the car starts from rest on level ground, and the power is constant, what is the speed of the car after 2.00 s and after 4.00 s? (Ignore any energy lost to friction.)𝑣

A car delivers 100. hp to the wheels. The mass of the car and driver is 1800. kg. If the car starts from rest on level ground, and the power is constant, what is the speed of the car after 2.00 s and after 4.00 s? (Ignore any energy lost to friction.)

Physics 115 - 04Linear MomentumDefine:Linear Momentum:For a particle of mass mmoving with velocity .SI Unit: kg∙m/s(no special name)Momentum is a vector which points in the same direction as the velocity.This definition turns out to be useful when analyzing the interactions between objects or particles.vvpm=

Physics 115 - 04Impulse-Momentum TheoremSuppose a constant net force acts on an object of mass mover a period of time t.=FFnetThen from Newton’s second law:aFm=nettm=v−=tmifvvtif−=ppUsing: vpm=Sonetfit =−= FpppI== Impulse

Physics 115 - 04Impulse-Momentum TheoremThis expression is the Impulse-Momentum Theorem.We derived this expression assuming that the net force is constant.pFI==tnet

Physics 115 - 04ImpulseIn general, a force is not constant during the time it acts.e.g.ivfvFtitfFirst contact with bat.Ball leaves battifttt−=Force on ball.First contact with bat.Ball leaves bate.g. Ball hitting a bat.

Physics 115 - 04ImpulseIf net force is not constant:But how do we actually calculate the impulse?In 1-dimension:()t=avImpulseFImpulse area under vs graph between and xifFttt=tt(We could prove this using calculus.)Average net force during contact.,avxFt=

Physics 115 - 04Impulse-Momentum TheoremOften is so small we can neglect the effects of some of the forces that act during t.e.g. We can neglect gravity in the baseball and bat collision and only consider the force of the bat on the ball.ifttt−=pF=tav

Physics 115 - 04SimilaritiesWork–Energy Theorem:Work = Force x Displacement= Change in Kinetic EnergyImpulse–Momentum Theorem:Impulse = Force x Time= Change in MomentumcosNetNetWFxKE== NetNett= = IFp(Scaler equation)(Vector equation)= Area under force vs. distance graph.= Area under force vs. time graph.

Lecture Example 32A ball of putty, with mass 0.20 kg is thrown towards a wall with a speed of 16 m/s. It sticks to the wall and the duration of the collision is 0.70 s. What is the impulse on the ball? What is the average force on the ball? Assume the positive direction is in the initial velocity direction. Note: the duration of the collision is the time from when the putty first makes contact with the wall to when it finally stops moving.

A ball of putty, with mass 0.20 kg is thrown towards a wall with a speed of 16 m/s. It sticks to the wall and the duration of the collision is 0.70 s. What is the impulse on the ball? What is the average force on the ball? Assume the positive direction is in the initial velocity direction.