Chapter 8: Potential Energy And Conservation Of Energy .

Chapter 8: Potential Energy andConservation of EnergyWork and kinetic energy are energies of motion.Consider a vertical spring oscillating with mass mattached to one end. At the extreme ends of travelthe kinetic energy is zero, but something caused itto accelerate back to the equilibrium point.We need to introduce an energy that depends onlocation or position. This energy is called potentialenergy.

Chapter 8:Potential Energy and Conservationof EnergyWork done by gravitation for a ball thrownupward that then falls back downbWab Wba - mgd mg d 0The gravitational force is said to be aconservative force.A force is a conservative force if the network it does on a particle moving aroundevery closed path is zero.a

Conservative forcesWab,1 Wba,2 0Wab,2 Wba,2 0therefore: Wab,1 Wab,2i.e. The work done by a conservative force on aparticle moving between two points does notdepend on the path taken by the particle.So, . choose the easiest path!!

Conservative & Non-conservative forcesConservative Forces: (path nservative forces: (dependent on path)frictionair resistanceelectrical resistanceThese forces will convert mechanical energy intoheat and/or deformation.

Gravitation Potential EnergyPotential energy is associated with the configuration of asystem in which a conservative force acts: U -WFor a general conservative force F F(x)Gravitational potential energy:assume Ui 0 at yi 0 (reference point)U(y) mgy(gravitational potential energy)only depends on vertical position

Nature only considers changes in Potential energyimportant. Thus, we will always measure U.So, . . . We need to set a reference point!The potential at that point could be defined to bezero (i.e., Uref. point 0) in which case we drop the“ ” for convenience.BUT, it is always understood to be there!

Potential energy and reference point (y 0 at ground)A 0.5 kg physics book falls from a table 1.0 m to the ground.What is U and U if we take reference point y 0 (assumeU 0 at y 0) at the ground?y U Uf - Ui -(mgh)Physicsy h0The book lost potential energy.Actually, it was converted intokinetic energyhy 0

Potential energy and reference point ( y 0 at table)A 0.5 kg physics book falls from a table 1.0 m to the ground.What is U and U if we take reference point y 0 (assumeU 0 at y 0) at the table?y U Uf - Ui mg(-h)Physicsy 00hThe book lost the same potentialenergy.(independent of y axis assignment)y -h

Sample problem 8-1: A block of cheese, m 2 kg, slidesalong frictionless track from a to b, the cheese traveled atotal distance of 2 m along the track, and a net verticaldistance of 0.8 m. How much is Wg? Fg drbWnetaBut, . We don’t know theangle between F and drEasier (or another) way?

Sample problem 8-1: A block of cheese, m 2 kg, slidesalong frictionless track from a to b, the cheese traveled atotal distance of 2 m along the track, and a net verticaldistance of 0.8 m. How much is Wg?cSplit the problem into two parts(two paths) since gravity isconservative0} dmg(b-c) mgd

Elastic Potential EnergySpring force is also a conservative forceF -kxUf – Ui ½ kxf2 – ½ kxi2Choose the free end of the relaxed spring as thereference point:that is: Ui 0 at xi 0U ½ kx2 Elastic potential energy

Conservation of Mechanical Energy Mechanical energyEmec K U For an isolated system (no external forces), if there areonly conservative forces causing energy transfer withinthe system .We know: K W(work-kinetic energy theorem)Also: U -W(definition of potential energy)Therefore: K U 0 (Kf – Ki) (Uf – Ui) 0therefore K1 U1 K2 U2 (States 1 and 2 of system)Emec,1 Emec,2 the mechanical energy is conserved

Conservation of mechanical energyFor an object moved by a spring in the presence of agravitational force.This is an isolated system with only conservativeforces ( F mg, F -kx ) acting inside the systemEmec,1 Emec,2K1 U1 K2 U2½ mv12 mgy1 ½ kx12 ½ mv22 mgy2 ½ kx22

The bowling ball pendulum demonstrationMechanicalenergy isconserved ifthere are onlyconservativeforces actingon the system.

Pendulum (1-D) verticallyFind the maximum speed of the mass.θIsolated system with only gravity(conservative force) acting on it.L13ymgy 0Let 1 be at maximum height (y ymax), 2Let 2 be at minimum height (y 0), and y L(1-cosθ)3 be somewhere between max and min.

Pendulum (1-D) verticallyIsolated system with only gravity(conservative force) acting on it.0K1 U1 K2 U2 Etotal2½ mv1 mgy1 ½0mv22 mgy20Kmax must be when Umin ( 0)qLcosθLWhere is the speed a max?3L1y2mgy 0In general,y L(1-cosθ)Answer: 2

Pendulum (1-D) verticallyFind the maximum speed of the mass.Let 1 be at maximum height (y ymax)and 2 be at minimum height (y 0)Emech,1 Emech,2 Etotaly L(1-cosθ)K1 U1 K2 U2 Etotaly00mgymax ½ mvmax2 Etotaly 0θLmg

Pendulum (1-D) verticallyNow find the speed of the mass as afunction of angle.Let 1 be at maximum height (y ymax)and 2 be at some height (y)Emech,1 Emech,2 Etotaly L(1-cosθ)K1 U1 K2 U2 Etotalmgymax ½ mv2 mgy EtotalθLyy 0mg

Pendulum (1-D) verticallyFind the speed of the mass as afunction of angle.Let 1 be at maximum height (y ymax)and 2 be at some height (y)θyy 0UK0-θmax0 θmax0-θmaxLθmg

Pendulum (1-D) verticallyFind the speed of the mass as afunction of angle.Let 1 be at maximum height (y ymax)and 2 be at some height (y)θyy 0Sum of U K is constant, EUEK0-θmax0 θmax0-θmaxLθmg

Potential Energy Curve We know U(x) -W -F(x) xTherefore F(x) -dU(x)/dx

Work done by external force When no friction acts within the system, the network done by the external force equals to thechange in mechanical energyW Emec K U Friction is a non-conservative force When a kinetic friction force acts within thesystem, then the thermal energy of the systemchanges: Eth fkdThereforeW Emec Eth

Work done by external force When there are non-conservative forces (likefriction) acting on the system, the net work doneby them equals to the change in mechanic energyWnet Emec K UWfriction -fkd

Conservation of EnergyThe total energy E of a system can change only byamounts of energy that are transferred to or from thesystemW E Emec Eth EintIf there is no change in internal energy, but friction actswithin the system: W Emec EthIf there are only conservative forces acting within thesystem: W EmecIf we take only the single object as the systemW K

Law of Conservation of Energy For an isolated system (W 0), the total energy Eof the system cannot change Emec Eth Eint 0 For an isolated system with only conservativeforces, Eth and Eint are both zero. Therefore: Emec 0

Chapter 8: Potential Energy andThe Conservation of Total EnergyWork and kinetic energy are energies of motion.Potential energy is an energy that depends on location.1-Dimension3-Dimensions

Force Potential Energy RelationshipF(x) -dU(x)/dxThus, the force in the x-direction is the negative derivative of thepotential energy! The same holds true for y- and z-directions.

Potential Energy CurveThus, what causes a force is the variation of the potential energyfunction, i.e., the force is the negative 3-D derivative of thepotential energy!

Potential Energy CurveWe know:Therefore: U(x) -W -F(x) xF(x) -dU(x)/dxNow integrate along the displacement: Rearrange terms:

Conservation of Mechanical EnergyHolds only for an isolated system (no external forces) andif only conservative forces are causing energy transferbetween kinetic and potential energies within the system.Mechanical energy:We know: K W U -WEmec K U(work-kinetic energy theorem)(definition of potential energy)Therefore: K U 0 Rearranging terms:(Kf – Ki) (Uf – Ui) 0Kf Uf Ki Ui K2 U2 Emec (a constant)

-Ax 0AHorizontal spring with mass oscillating with maximumamplitude xmax A. At which displacement(s) wouldthe kinetic energy equal the potential energy?5) none of the above

Uf – Ui ½ kxf2 – ½ kxi2Ui 0 at xi 0; Umax ½ kxmax2Since K UHorizontal spring with mass oscillating withmaximum amplitude xmax A. At whichdisplacement(s) would the kinetic energy equalthe potential energy?5) none of the above-Ax 0A

Conservation of mechanical energyFor an isolated system with only conservative forces (e.g.,F mg and F -kx) acting on the system:Emec,1 Emec,2 Etotal K1 U1 K2 U2 Etotal½ mv12 mgx1 ½ kx12 ½ mv22 mgx2 ½ kx22 EtotalInitial mechanical energyFinal mechanical energy

Horizontal Springx 0Isolated system with only conservative forces acting on it.(e.g.,)Emech,1 Emech,2 EtotalK1 U1 K2 U2 Etotal½ mv12 ½ kx12 ½ mv22 ½ kx22 Etotal

Work done by Spring ForceSpring force is a conservative forceWork done by the spring force:If xf xi (further away from equilibrium position); Ws 0My hand did positive work, while the spring did negativework so the total work on the object 0

Work done by Spring ForceSpring force is a conservative forceWork done by the spring force:If xf xi (closer to equilibrium position); Ws 0My hand did negative work, while the spring did positivework so the total work on the object 0

Work done by Spring Force -- SummarySpring force is a conservative forceWork done by the spring force:If xf xi (further away from equilibrium position); Ws 0If xf xi (closer to equilibrium position); Ws 0Let xi 0, xf xthen Ws - ½ k x2

Elastic Potential EnergySpring force is a conservative forceChoose the free end of the relaxed spring as the referencepoint: Ui 0 at xi 0The work went into potential energy, since thespeeds are zero before and after.

Vertical Spring with mass mConsider that mass m was held so thespring was relaxed (y1 0) and then slowly let down to the equilibrium position.y1 0Find this equilibrium position.y -y0Equilibrium position (Fg Fs 0):Fg -mgFs -ky0(now the spring is not relaxed!)y2 -A y0 -mg/k (substitute for yo)Ke 0, Ug,e mgy0 mg(-mg/k) -(mg)2/k,Us,e ½ ky02 ½ k (-mg/k)2 ½ (mg)2/k-kymg

Vertical Spring with mass mNow consider that the mass m wasdropped from y1 0Find maximum amplitude A. y1 0Initial position (y1 0):y -y02K1 0, Ug,1 0, Us,1 ½ ky 0Since y1 0Choose 0 Emech,1 0y2 -AFinal position (y2 A) :K2 0, Ug,2 mgy2 mg(-A) -mgA,Us,2 ½ ky22 ½ kA2Emech,1 Emech,2 0 Ug,2 Us,2 -mgA ½ kA2 A 2mg/k( 2y0)-kymg

Vertical Spring with mass mNow consider that the mass m wasdropped from y1 0Where does the max speed occur?Maximum speed occurs when thepotential energy is a minimum. y1 0y -y0For an Arbitrary position:-kymgy2 -AEmech 0 K Ug Us ½ mv2 mgy ½ ky2dU/dy d(mgy ½ ky2)/dy 0 mg ky 0 y -mg/k which is yo (from 2 slides ago) maximum speed is at equilibrium point (as expected)

Vertical Spring with mass mNow consider that the mass m wasdropped from y1 0What is the max speed?Maximum speed position: y1 0y -y0-kymg0 Kmax Ug Us½ mv2 mg(-mg/k) ½ k(-mg/k)2½ mv2 - (mg)2/k ½ k(mg/k)2½ mv2 - ½ (mg)2/k v2 mg2/ky2 -A

Vertical Spring with mass mNow consider that the mass m was dropped from y1 0Find the mechanical energy of the mass at y 01the equilibrium point.-kyy -y0Kinetic energy at y -y0:Ky0 ½ (mg)2/kmgy2 -APotential energy at y - y0:(Ug Us) y -y0 mg(-y0) ½ k(-y0)2 -mgy0 ½k(y0)2 -mg(mg/k) ½k(mg/k)2 -½(mg)2/kMechanical energy at y - y0:Emech Ky0 (Ug Us) y -y0 ½ (mg)2/k -½(mg)2/k 0

Vertical Spring with mass mWhen we slowly let the mass relax to -y0, the mechanical energy is:y1 0E K (U U ) mech-y0gs y -y00-(mg)2/k ½ (mg)2/k Emech -½(mg)2/kBut when we let the mass drop, themechanical energy at y y0 is Emech 0y -y0-kymgy2 -AWhere did the missing energy (½(mg)2/k) go?Answer: Our hand provided a non-conservative force on m!

Work done by external forceWhen no friction acts within the system, the net workdone by an external force from outside the systemequals the change in mechanical energy of the systemWnet Emec K UFriction is a non-conservative force that opposes motionWork done by friction is: Wfriction -fkdWhen a kinetic friction force acts within the system,then the thermal energy of the system changes: Eth fkdThereforeW Emec Eth