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CHAPTER 8- POTENTIAL ENERGY and CONSERVATION of ENERGY POTENTIAL ENERGY Potential Emergy is energy stored by thepositioning of an object in a force field or by the positioning of a mechanical object. y U = m g (y-yo) Fg gravity dy yo U = 1/2 k ∆x2 spring Fs ∆x dx In each case an increase in potential energy ∆U is associated with an amount of work W=- ∫ F dx where the negative sign implies the force and displacement are in opposite directions. CONSERVATION OF ENERGY Since a positive change in kinetic energy ∆K is associated with positive work we can say potential and kinetic energy work oppositely, that is W = ∆K = - ∆U or ∆K + ∆U = 0 If we define an objects Total Energy as E = K+ U then ∆E = 0 along the path of any object acted upon by a force. Any loss of energy due to frictional forces Wf must be included. EB EA Wf EA = EB + | Wf | ∆EBA = EB - EA = | Wf | SUPER MARIO (No Friction) If a spring-mass (Super Mario in class ) is compressed by ∆y and released how high will it jump above the table? E(bot) = 1/2 k ∆y2 (starting from rest K = 0) E(top) = m g h (At top K = 0) Since energy is conserved E(bot) = E(top) 1/2 k ∆y2 = m g h h = ( 1/2 k ∆y2 )/ mg ∆y h INCLINED PLANE A mass M slides down an inclined plane from a height H. What is its velocity at the bottom? From conservation of energy over the frictionless surface we can write H E(bot) = E(top) 1/2 m V2 = m g H V= (2gH) θ V 1/2 INCLINED PLANE with friction A mass M slides down an inclined plane (coeficient of kinetic friction µK) from a height H. What is its velocity at the bottom? From conservation of energy over the frictionless surface we can write H E(bot) = E(top) - |Wf | = E(top) - µ M g sin(θ) L 1/2 M V2 = M g H - µ M g sin(θ) (H / sin(θ) ) V = ( 2 g H - 2 µ g H tan(θ) ) 1/2 L θ V LOOP-THE-LOOP A mass M ris eleased from a height 4R and enters a circular loop-the-loop of radius R. EA = M g 4R EB = 1/2 M VB2 EC = 1/2 M VC2 + M g R A ED = 1/2 M VD2 + M g 2R MV2/R N g Conservation of energy EA = EB = EC = ED Mg D 4R C B Does the mass M stay on the track at D or fall down? Force balance at D gives: N + Mg =MVD2 / R If N=0 then the mass may lift-off. Mg < MVD2 / R is the condition for falling such that VD is not large enough. GLOBAL ENERGY CONSERVATION Even though we defined cases where energy was not conserved due to frictional forces for example, if we could account for theenergy loss we could make the statement that “global energy” or “total energy” was always conserved! Scientist believe that the “total energy” of the universe will remain the same even though it may change its form from electrical, mechanical, gravity, frictional, heat, atomic, nuclear, etc. E = K + Ue + Umech + Ug + Ufrict + Uheat + Uatomic + Unucl = Universal Constant EINSTEIN’S ENERGY Einstein defined energy in relativistic terms. He counted the mass Mo of an object as part of its energy. This shocked the world of classical physics! E = M c2 = K + U + Mo c2 He realized that if nucleus A split in to B and C and MA > MB + MC or ∆M= MA –( MB + MC) and excess of energy is released E = ∆M c2 This excess of energy comes from the nuclear binding potential Unuc If a uranium atom releases E=10-10 J and there are 6.02 x 1021 atoms in a mole then Etot = 6 x 1011 J of energy released !