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Chapter 8C - Conservation of Energy A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State 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 Energy is the ability to do work by virtue of position or 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 Potential Energy U is equal to the work that can be done BY gravity due to height above a specified point. U = 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) U = 1960 J The Origin of Potential Energy Potential energy is a property of the Earthbody system. Neither has potential energy without the other. F mg h Work done by lifting force F provides positive potential energy, mgh, for earthbody system. Only external forces can add or remove energy. Conservative Forces A conservative force is one that does zero work during a round trip. F mg h Weight is conservative. Work done by earth on the way up is negative, - mgh Work on return is positive, +mgh Net Work = - mgh + mgh = 0 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 work = 0 (conservative) x m F Independence of Path Work done by conservative forces is independent of the path. C A B Force due to gravity mg C A B Because vertical of Work (A only C) =the Work (A B component C) Why? the weight does work against gravity. Nonconservative Forces Work done by nonconservative forces cannot be restored. Energy is lost and cannot be regained. It is 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 work done against friction is greater for the longer path (BCD). Stored Potential Energy Work done by a conservative force is stored in the system as potential energy. m x xo The potential energy is equal to the work done in compressing the 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 the absence of friction, the sum of the potential and kinetic energies is a constant, provided no energy is added to 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 E = U + K = 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 Etop = 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 v = 17.1 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 v = 80 ft/s Conservation of Energy in Absence of Friction Forces The total energy is constant for a conservative system, such as with gravity or a 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 v2f 2 gh (25.8 m/s)2 2(9.8 m/s2 )(33.2 m) v0 14.9 m /s 2 2 vo = 3.86 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 2gh (10 m/s) 2(9.8 m/s )(4 m) 2 f 2 0 v f 21.6 m2 /s2 2 2 vf = 4.65 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 ½kxo 2 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 h = = 0.327 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 sin 30o = s= h sin 30o End Begin s 30o h s = 32.7 cm Sin 30o s = 65.3 cm h Energy Conservation and Nonconservative Forces. f Work against friction forces must be accounted for. Energy is still conserved, but not reversible. Conservation of Mechanical Energy (U + K)o = (U + K)f + Losses 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? U = mgh b. Do I have velocity? K = ½mv2 c. Do I have a spring? U = ½kx2 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) vc = 8.85 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 mk = 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 = (mkn) x = mk(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 mk = 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 = mk(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 mk = 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 v =11.4 m/s Summary: Energy Gains or Losses: Gravitational Potential Energy U = mgh Spring Potential Energy U kx Kinetic Energy K mv Work Against Friction 1 2 1 2 2 2 Work = 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