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Chapter 8C - Conservation of Energy AA PowerPoint PowerPoint Presentation Presentation by by Paul Paul E. E. Tippens, Tippens, Professor Professor of of Physics Physics Southern Southern Polytechnic Polytechnic State State University University © 2007 A waterfall in Yellowstone Park provides an example of energy in nature. The potential energy of the water at the top is converted into kinetic energy at the bottom. Objectives: After completing this module, you should be able to: • Define and give examples of conservative and nonconservative forces. • Define and apply the concept of conservation of mechanical energy for conservative forces. • Define and apply the concept of conservation of mechanical energy accounting for friction losses. Potential Energy Potential Potential Energy Energy isis the the ability ability to to do do work work by by virtue virtue of of position position or or condition. condition. h m Example: A mass held a distance h above the earth. mg If released, the earth can do work on the mass: Earth Work = mgh Is this Positive! work + or - ? Gravitational Potential Energy Gravitational Gravitational Potential Potential Energy Energy UU isis equal equal to to the the work work that that can can be be done done BY BY gravity gravity due due to to height height above above aa specified specified point. point. UU = = mgh mgh Gravitational P. E. Example: What is the potential energy when a 10 kg block is held 20 m above the street? U = mgh = (10 kg)(9.8 m/s2)(20 m) UU == 1960 1960 JJ The Origin of Potential Energy Potential Potential energy energy isis aa property property of of the the Earth Earthbody body system. system. Neither Neither has has potential potential energy energy without without the the other. other. Work done by F lifting force F h provides positive mg potential energy, mgh, for earthbody system. Only external forces can add or remove energy. energy Conservative Forces AA conservative conservative force force isis one one that that does does zero zero work work during during aa round round trip. trip. Weight is conservative. F Work done by earth h on the way up is mg negative, - mgh Work on return is positive, +mgh Net Net Work Work = = -- mgh mgh + + mgh mgh = = 00 The Spring Force The force exerted by a spring is also conservative. F x m When stretched, the spring does negative work, - ½kx2. On release, the spring does positive work, + ½kx2 Net Net work work == 00 (conservative) (conservative) x m F Independence of Path A Work Work done done by by conservative conservative forces forces isis independent independent of of the the path. path. C C Force due to gravity mg B B A Because vertical of Work (A only C) =the Work (A Bcomponent C) Why? the weight does work against gravity. Nonconservative Forces Work Work done done by by nonconservative nonconservative forces forces cannot cannot be be restored. restored. Energy Energy isis lost lost and and cannot -dependent! cannot be be regained. regained. ItIt isis path path-dependent! B A f A B m f Friction forces are nonconservative forces. Work of Conservative Forces is Independent of Path: For gravitational force: B C (Work)AB= -(Work)BCA Zero net work For friction force: A (Work)AB -(Work)BCA The The work work done done against against friction friction isis greater greater for for the the longer longer path path (BCD). (BCD). Stored Potential Energy Work done by a conservative force is stored in the system as potential energy. m x xo The The potential potential energy energy isis equal equal to to the the work work done done in in compressing compressing the the spring: spring: F(x) = kx to compress Potential energy of compressed spring: Displacement is x U Work kx 1 2 2 Conservation of Energy (Conservative forces) In In the the absence absence of of friction, friction, the the sum sum of of the the potential potential and and kinetic kinetic energies energies isis aa constant, constant, provided provided no no energy energy isis added added to to system. system. h y v=0 v mg At top: Uo = mgh; Ko = 0 At y: Uo = mgy; Ko = ½mv2 At y=0: Uo = 0; Ko = ½mvf 2 0 vf EE == UU ++ KK == Constant Constant Constant Total Energy for a Falling Body TOP: E = U + K = mgh h K=0 y At any y: E = mgh + ½mv2 v Bottom: E = ½mv2 mgh = mgy + ½mv2 = ½mvf2 Total E is same at any point. (Neglecting Air Friction) 0 U=0 vf Example 1: A 2-kg ball is released from a height of 20 m. What is its velocity when its height has decreased to 5 m? 20m Total = Total E at 5 m Total EEtop top = Total E at 5 m v=0 mgh = mgy + ½mv2 2gh = 2gy + v2 5m v2 = 2g(h - y) = 2(9.8)(20 - 5) v= (2)(9.8)(15) 0 v vv == 17.1 17.1 m/s m/s Example 2: A roller coaster boasts a maximum height of 100 ft. What is the speed when it reaches its lowest point? Assume zero friction: At top: U + K = mgh + 0 Bottom: U + K = 0 + ½mv2 Total energy is conserved mgh = ½mv2 v= (2)(32 ft/s2)(100 ft) v = 2gh vv = = 80 80 ft/s ft/s Conservation of Energy in Absence of Friction Forces The The total total energy energy isis constant constant for for aa conservative conservative system, system, such such as as with with gravity gravity or or aa spring. spring. Begin: (U + K)o = End: (U + K)f Height? mgho Spring? ½kxo2 Velocity? ½mvo2 = mghf Height? ½kxf2 Spring? ½mvf2 Velocity? Example 3. Water at the bottom of a falls has a velocity of 30 m/s after falling 35 ft. What is the water speed at the top of the falls? ho = 35 m; vf = 30 m/s2 First look at beginning point—top of falls. Assume y = 0 at bottom for reference point. Height? Yes (35 m) mgho Spring? No ½kxo2 Velocity? Yes (vo) ½mvo2 Example 3 (Cont.) Water at the bottom of falls has a velocity of 30 m/s after falling 35 ft. What is the water speed at the top of the falls? ho = 35 m; vf = 30 m/s2 Next choose END point at bottom of falls: Height? No (0 m) mghf Spring? No ½kxf2 Velocity? Yes (vf) ½mvf2 Example 3 (Cont.) Water at the bottom of falls has a velocity of 30 m/s after falling 35 ft. ho = 35 m; vf = 30 m/s2 What is the water speed at the top of the falls? Total energy at top = Total energy at bottom mgh mv 0 mv 1 2 2 0 1 2 2 f 2 gh v v 2 0 2 f v02 v 2f 2 gh (25.8 m/s) 2 2(9.8 m/s 2 )(33.2 m) v0 14.9 m /s 2 2 vvoo== 3.86 3.86 m/s m/s Example 4. A bicycle with initial velocity 10 m/s coasts to a net height of 4 m. What is the velocity at the top, neglecting friction? E(Top) = E(Bottom) vf = ? 4m 1 2 Etop = mgh + ½mv2 vo = 10 m/s EBot = 0 + ½mvo2 mv mgh mv 2 f 1 2 2 0 1 2 v v gh 2 f 1 2 2 0 v v 2 gh (10 m/s) 2(9.8 m/s )(4 m) 2 f 2 0 v f 21.6 m 2 /s 2 2 2 vvff == 4.65 4.65 m/s m/s Example 5: How far up the 30o-incline will the 2-kg block move after release? The spring constant is 2000 N/m and it is compressed by 8 cm. mgho ½kxo2 ½mvo2 = mghf End Begin ½kxf2 s 30o h ½mvf2 Conservation of Energy: kx02 (2000 N/m)(0.08m) 2 h 2 2mg 2(2 kg)(9.8 m/s ) ½kxo2 = mghf hh == == 0.327 0.327 m m Example (Cont.): How far up the 30oincline will the 2-kg block move after release? The spring constant is 2000 N/m and it is compressed by 8 cm. Continued: h = 0.327 m = 32.7 cm h sin 30o = s h 32.7 cm s= = sin 30o Sin 30o End Begin s 30o ss == 65.3 65.3 cm cm h Energy Conservation and Nonconservative Forces. f Work Work against against friction friction forces forces must must be be accounted accounted for. for. Energy Energy isis still still conserved, conserved, but but not not reversible. reversible. Conservation of Mechanical Energy (U = (U (U + + K) K)ff + + Losses Losses (U + + K) K)oo = Problem Solving Strategies 1. Read the problem; draw and label a sketch. 2. Determine the reference points for gravitational and/or spring potential energies. 3. Select a beginning point and an ending point and ask three questions at each point: a. Do I have height? UU == mgh mgh b. Do I have velocity? 22 KK == ½mv ½mv c. Do I have a spring? 22 UU == ½kx ½kx Problem Solving (Continued) 4. Apply the rule for Conservation of Energy. mgho ½kxo2 ½mvo2 = mghf ½kxf2 ½mvf2 + Work against friction: fk x 5. Remember to use the absolute (+) value of the work of friction. (Loss of energy) Example 6: A mass m is connected to a cord of length L and held horizontally as shown. What will be the velocity at point B? (d = 12 m, L = 20 m) A 1. Draw & label. L v B c 2. Begin A and end B. d r 3. Reference U = 0. 0 (U + K)o =(U + K)f + loss mgL + 0 = mg(2r) + ½mvc2 2gL - 4gr = vc2 U=0 (Multiply by 2, simplify) Next find r from figure. Example (Cont.): A mass m is connected to a cord of length L and held horizontally as shown. What will be the velocity at point B? (d = 12 m, L = 20 m) A 2gL - 4gr = vc2 L vc r=L-d r = 20 m - 12 m = 8 m vc2 =2gL - 4gr = 2g(L - 2r) B r d U=0 vc2 = 2(9.8 m/s2)[20 m - (2)(8 m)] vc = 2(9.8 m/s2)(4 m) vvcc == 8.85 8.85 m/s m/s Example 7: A 2-kg mass m located 10 m above the ground compresses a spring 6 cm. The spring constant is 40,000 N/m and k = 0.4. What is the speed when it reaches the bottom? 2 kg h s 30o n f Begin mg Cos End mg Sin 30o 30o 30o mg Conservation: mgh + ½kx2 = ½mv2 + fkx (Work)f = (kn) x = (mg Cos 30o) x Continued . . . Example (Cont.): A 2-kg mass m located 10 m above the ground compresses a spring 6 cm. The spring constant is 40,000 N/m and k = 0.4. What is the speed when it reaches the bottom? mgh + ½kx2 = ½mv2 + fkx 2 kg h 10 m x 30o x= 10 m = 20 m Sin 30o fkx = (mg Cos 30o) x fkx = (0.4)(2 kg)(9.8 m/s2)(0.866)(20 m) = 136 J mgh = (2 kg)(9.8 m/s2)(10 m) = 196 J ½kx2 = ½(40,000 N/m)(0.06 m)2 = 72.0 J Example (Cont.): A 2-kg mass m located 10 m above the ground compresses a spring 6 cm. The spring constant is 40,000 N/m and k = 0.4. What is the speed when it reaches the bottom? mgh + ½kx2 = ½mv2 + fkx 2 kg h 10 m x 30o mgh = 196 J ½kx2 = 72.0 J fkx = 136 J ½mv2 = mgh + ½kx2 - fkx ½(2 kg) v2 = 196 J + 72 J - 136 J = 132 J vv =11.4 =11.4 m/s m/s Summary: Energy Gains or Losses: Gravitational Gravitational Potential Potential Energy Energy UU == mgh mgh Spring Spring Potential Potential Energy Energy U kx Kinetic Kinetic Energy Energy K mv Work Work Against Against Friction Friction 1 2 1 2 2 2 Work Work = = fx fx Summary: Conservation of Energy The basic rule for conservation of energy: mgho ½kxo2 ½mvo2 = mghf ½kxf2 ½mvf2 + Work against friction: fk x Remember to use the absolute (+) value of the work of friction. (Loss of energy) CONCLUSION: Chapter 8C Conservation of Energy