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.

If released, the earth can do work on the mass: Potential Energy Potential Energy is the ability to do work by virtue of position or condition. Example: A mass held a distance h above the earth. Earth mg h m If released, the earth can do work on the mass: Work = mgh Is this work + or - ? Positive!

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 Earth-body system. Neither has potential energy without the other. Work done by lifting force F provides positive potential energy, mgh, for earth-body system. mg h F Only external forces can add or remove energy.

Weight is conservative. Conservative Forces A conservative force is one that does zero work during a round trip. Weight is conservative. mg h F 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 F x m The force exerted by a spring is also conservative. When stretched, the spring does negative work, - ½kx2. F x m On release, the spring does positive work, + ½kx2 Net work = 0 (conservative)

Work done by conservative forces is independent of the path. Independence of Path Work done by conservative forces is independent of the path. A C B C A B Force due to gravity mg Because only the vertical component of the weight does work against gravity. Work (A C) = Work (A B C) Why?

Nonconservative Forces Work done by nonconservative forces cannot be restored. Energy is lost and cannot be regained. It is path-dependent! B A f f m A B Friction forces are nonconservative forces.

Work of Conservative Forces is Independent of Path: B C For gravitational force: (Work)AB= -(Work)BCA Zero net work For friction force: (Work)AB ¹ -(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 xo x The potential energy is equal to the work done in compressing the spring: F(x) = kx to compress Displacement is x Potential energy of compressed spring:

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. v = 0 h At top: Uo = mgh; Ko = 0 mg v y At y: Uo = mgy; Ko = ½mv2 At y=0: Uo = 0; Ko = ½mvf 2 vf E = U + K = Constant

Constant Total Energy for a Falling Body K = 0 h TOP: E = U + K = mgh v y At any y: E = mgh + ½mv2 Bottom: E = ½mv2 mgh = mgy + ½mv2 = ½mvf2 U = 0 vf Total E is same at any point. (Neglecting Air Friction)

Example 1: A 2-kg ball is released from a height of 20 m 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? v = 0 20m Total Etop = Total E at 5 m mgh = mgy + ½mv2 v 5m 2gh = 2gy + v2 v2 = 2g(h - y) = 2(9.8)(20 - 5) v = 17.1 m/s v = (2)(9.8)(15)

Total energy is conserved 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 = 2gh v = 80 ft/s v = (2)(32 ft/s2)(100 ft)

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? Spring? Velocity? mgho ½kxo2 ½mvo2 = mghf ½kxf2 ½mvf2 Height? Spring? Velocity?

What is the water speed at the top of the falls? Example 3. Water at the bottom of a 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? First look at beginning point—top of falls. Assume y = 0 at bottom for reference point. Height? Spring? Velocity? Yes (35 m) mgho ½kxo2 ½mvo2 No Yes (vo)

Next choose END point at bottom of falls: 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? Next choose END point at bottom of falls: Height? Spring? Velocity? No (0 m) mghf ½kxf2 ½mvf2 No Yes (vf)

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 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? 4 m vf = ? vo = 10 m/s E(Top) = E(Bottom) Etop = mgh + ½mv2 EBot = 0 + ½mvo2 vf = 4.65 m/s

Conservation of Energy: 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. End mgho ½kxo2 ½mvo2 = mghf ½kxf2 ½mvf2 s h 30o Begin Conservation of Energy: ½kxo2 = mghf h = = 0.327 m

Example (Cont.): 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. s h 30o Begin End Continued: h = 0.327 m = 32.7 cm sin 30o = h s s = = h sin 30o 32.7 cm Sin 30o s = 65.3 cm

Energy Conservation and Nonconservative Forces. Work against friction forces must be accounted for. Energy is still conserved, but not reversible. f 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 gravi- tational 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)

(Multiply by 2, simplify) 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 B L vc r 1. Draw & label. d 2. Begin A and end B. 3. Reference U = 0. (U + K)o =(U + K)f + loss U = 0 mgL + 0 = mg(2r) + ½mvc2 (Multiply by 2, simplify) 2gL - 4gr = vc2 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) B L vc r d U = 0 A 2gL - 4gr = vc2 r = L - d r = 20 m - 12 m = 8 m vc2 =2gL - 4gr = 2g(L - 2r) vc2 = 2(9.8 m/s2)[20 m - (2)(8 m)] vc = 8.85 m/s vc = 2(9.8 m/s2)(4 m)

(Work)f = (mkn) x = mk(mg Cos 30o) x 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? h 2 kg s 30o mg f n mg Sin 30o mg Cos 30o 30o Begin End Conservation: mgh + ½kx2 = ½mv2 + fkx (Work)f = (mkn) x = mk(mg Cos 30o) x Continued . . .

fkx = (0.4)(2 kg)(9.8 m/s2)(0.866)(20 m) = 136 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? h 2 kg x 30o 10 m mgh + ½kx2 = ½mv2 + fkx x = = 20 m 10 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? h 2 kg x 30o 10 m mgh + ½kx2 = ½mv2 + fkx fkx = 136 J mgh = 196 J ½kx2 = 72.0 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 Kinetic Energy Work Against Friction 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