Chapter 8 – Conservation of energyI. Work done on a system by an external forceII. Conservation of mechanical energyII. External work and thermal energyIII. External forces and internal energy changesIV. PowerI. Work done on a system by an external forceNo Friction:dvvaadvvmafFk/)(5.02202202Work is energy transfer “to” or “from” a system by means of an externalforce acting on that system.When more than one force acts on a system their network is the energy transferred to or from the system.W=∆Emec= ∆K+ ∆U  Ext. forceFriction:Remember! ∆Emec= ∆K+ ∆U=0only when:- System isolated.- No ext. forces act on a system.- All internal forces are conservative.dfKFdWdfmvmvFdvvmdfFdvvdmfFkkkk2022022022121)(21)(2dfEFdWkmecGeneral:Example: Block sliding up a ramp.dfEkthThermal energy:thmecEEFdWWork done on a system by an external force, friction involvedFriction due to cold welding between twosurfaces. As the block slides over the floor,the sliding causes tearing and reforming of thewelds between the block and the floor, whichmakes the block-floor warmer.II. Conservation of energy- The total energy of a system can only change by amounts of energytransferred “from” or “to” the system.intEEEEWthmec Experimental law-The total energy of an isolated system cannot change. (There cannot beenergy transfers to or from it).Isolated system:0intEEEthmecIn an isolated system we can relate the total energy at one instant to thetotal energy at another instant without considering the energies atintermediate states.Total energy of a system = E mechanical + E thermal + E internalExample: Trolley pole jumper1) Run  Internal energy (muscles)gets transferred into kinetic energy.2) Jump/Ascent  Kinetic energytransferred to potential elasticenergy (trolley pole deformation)and to gravitational potential energy3) Descent Gravitational potentialenergy gets transferred into kineticenergy.III. External forces and internal energy changesExample: skater pushes herself away from a railing. There is a force Fon her from the railing that increases her kinetic energy.i) One part of an object (skater’s arm) does not move like the rest of body.ii) Internal energy transfer (from one part of the system to another) via the external force F. Biochemical energy from muscles transferred to kinetic energy of the body.coscos)(cos,,FdEFdWUKsystemisolatedNondFKWmecextFextFChange in system’s mechanical energy by an external forceIV. PowertEPavgAverage power:Instantaneous power:dtdEP Proof:dFKdMaMvMvMdavvxx)cos(2121)5.0(220220261. In the figure below, a block slides along a path that is without friction until theblock reaches the section of length L=0.75m, which begins at height h=2m. In thatsection, the coefficient of kinetic friction is 0.4. The block passes through point Awith a speed of 8m/s. Does it reach point B (where the section of friction ends)? Ifso, what is the speed there and if not, what greatest height above point A does itreach?mgNfsmvmghmvmvUKUKEforcesveconservatiOnlyCAmmNfmmgNcccACCAAmeckk/5212104.3)5.8)(4.0(5.830cos22The kinetic energy in C turns into thermal and potential energy  Block stops.BreachesBlockmLdmetersdmddmgmdfmgyKmmvKkccc75.049.14.3)30sin(4.124.125.02smvmmmvmmgLmgLmvmgLyymgmvmLfUKUKEUEEsystemIsolatedBBkBkcBBkBBCCthmec/5.35.267.35.04.1230cos30sin5.030cos)(5.04.120222C129. A massless rigid rod of length L has a ball of mass m attached to one end. The other end is pivoted insuch a way that the ball will move in a vertical circle. First, assume that there is no friction at the pivot.The system is launched downward from the horizontal position A with initial speed v0. The ball justbarely reaches point D and then stops. (a) Derive an expression for v0in terms of L, m and g. (b) What isthe tension in the rod when the ball passes through B? (c) A little girl is placed on the pivot to increasethe friction there. Then the ball just barely reaches C when launched from A with the same speed asbefore. What is the decrease in mechanical energy during this motion? (d) What is the decrease in mecha-nical energy by the time the ball finally comes to rest at B after several oscillations?gLvmvmgLUKUKUKEaADiiffmec2210;00)(020DCABLv0yxmgTgLvvgLgLmvmgLmvKUKUgvLmTmgTLvmmgTmaFbBBBBBAABBccent522122121211)(222022mgTFcdfEEEEWvckththmecc 0)(The difference in heights or in gravitational potential energies between the positionsC (reached by the ball when there is friction) and D during the frictionless movementIs going to be the loss of mechanical energy which goes into thermal energy.mgLEcth)(mgLEmec2(d) The difference in height between B and D is 2L. The total loss of mechanical energy(which all goes into thermal energy) is:101. A 3kg sloth hangs 3m above the ground. (a) What is the gravitationalpotential energy of the sloth-Earth system if we take the reference point y=0 to beat the ground? If the sloth drops to the ground and air drag on it is assumed to benegligible, what are (b) the kinetic energy and (c) the speed of the sloth just beforeit reaches the ground?JmsmkgmghUsmmKvmvKcKgroundUJKbUKUKEaiffffiffiiffmec1.94)3)(/8.9)(2.3(/67.7221)(0;0)(1.94)(0)(22130. A metal tool is sharpen by being held against the rim of a wheel on a grindingmachine by a force of 180N. The frictional forces between the rim and the toolgrind small pieces of the tool. The wheel has a radius of 20cm and rotates at 2.5rev/s. The coefficient of kinetic friction between the wheel and the tool is 0.32. Atwhat rate is energy being transferred from the motor driving the wheel and thetool to the kinetic energy of the material thrown from the tool?F=180NvmotorpliedPowerfrictionbydissipatedPowerNNFNfsmrevmsrevvkkksup6.57)180)(32.0(/14.31)2.0(25.2WPWsmNvfPmotor181181)/14.3)(6.57(82. A block with a kinetic energy