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Physics 2053C – Fall 2001

Physics 2053C – Fall 2001. Chapter 6 Conservation of Energy Work & Energy. Conservation of Energy. Energy has a variety of forms Energy of Motion Kinetic Energy = ½ mv 2 Energy due to location or configuration Potential Energy mgh (gravitational potential energy)

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Physics 2053C – Fall 2001

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  1. Physics 2053C – Fall 2001 Chapter 6 Conservation of Energy Work & Energy Dr. Larry Dennis, FSU Department of Physics

  2. Conservation of Energy • Energy has a variety of forms • Energy of Motion • Kinetic Energy = ½ mv2 • Energy due to location or configuration • Potential Energy • mgh (gravitational potential energy) • ½ kx2 (potential energy in a spring) • Heat Energy (Random motion of atoms) • Generalized Work-Energy Theorem K1 + U1 + Wnc = K2 + U2

  3. Conservation of Energy Kinetic Energy Potential Energy When is K + U = constant? When is K + U not constant? Conservative forces Non-conservative forces Work done by friction

  4. Example Problem: Ms. Daredevil. A B Which path, A or B, produces the fastest speed at the bottom – assuming no friction?

  5. Example Problem: Ms. Daredevil. Where is our fearless daredevil’s potential energy the highest? Where is our fearless daredevil’s kinetic energy the highest? Where is our fearless daredevil’s total energy the highest?

  6. 10 m 21 m Example Problem: Ms. Daredevil. If the daredevil’s mass is 34 kg, how fast is she traveling at the bottom of this slope, assuming she starts from rest, does not push with her poles and there is no friction? K1 + U1 = K2 + U2 0 + mgh = ½mV2 + 0 2gh = V2 V = (2*9.8 m/s2*10 m)½ V = 14 m/s

  7. 10 m 21 m Example Problem: Ms. Daredevil. Now assume friction does 500 J of work on our fearless daredevil as she hurtles down the slope. What is her speed at the bottom? K1 + U1 + Wf = K2 + U2 0 + mgh + Wf = ½mV2 + 0 2(gh + Wf /m) = V2 V = (2*9.8 m/s2*10 m + 2*(-500)/34)½ V = 12.9 m/s

  8. 10 m 21 m Example Problem: Ms. Daredevil. Assuming friction does 500 J of work on our fearless daredevil as she hurtles down the slope, and the she travels 26 m in the process. What is the average force of friction between Ms. Daredevil and the slope? Wf = f*d f = Wf /d f = 500/26 = 19.2 N

  9. Sample Problems for Quiz 6 • Sample Questions: • Chap 6: 12, 18, 19 • Sample Problems: • Chap 6: 7, 35, 51, 75

  10. CAPA # 3 • A 32 kg child slides down a long slide in a playground. She starts from rest at a height of 20 m. When she is partway down the slide, at a height of 12 m, she is moving at a speed of 8.3 m/s. Calculate the mechanical energy lost due to friction. 20 m K1 + U1 + Wf = K2 + U2 0 + mgh1 + Wf = ½mv2 + mgh2 Wf = mgh2 - mgh1 + ½mv2 12 m

  11. CAPA # 3 • A 32 kg child slides down a long slide in a playground. She starts from rest at a height of 20 m. When she is partway down the slide, at a height of 12 m, she is moving at a speed of 8.3 m/s. Calculate the mechanical energy lost due to friction. 20 m Wf = mgh2 - mgh1 + ½mv2 Wf = 32*9.8*(12 – 20) + ½32*8.32 Wf = - 1407 J Energy lost is 1407 J. 12 m

  12. Loop-the-Loop  CAPA #7 & 8 Conserve Energy: K1 + U1 = K2 + U2 0 + mgh = ½mV2 + mg(2R) mg(h-2R) = ½mV2 2g(h-2R) = V2 V= (2g(h-2R))1/2 P A

  13. Loop-the-Loop  CAPA #7 & 8 Uniform Circular Motion: V= (2g(h-2R))1/2 a = V2 / R a= (2g(h-2R)) / R a = 2g * (h-2R)/R P A

  14. Power Power = Work/Time P = W/t = F*X/t = F * V Units are J/s or Watts (W)

  15. CAPA # 9: Power & Units of Energy The human body converts energy into work and heat at rates of 60 to 125 W (called the basal metabolic rate). This energy comes from food and is usually measured in kilocalories ( 1 kcal = 4.186 kJ). How many kilocalories of food energy does a person with a metabolic rate of 98.0 W require per day? (Assume 100% efficiency.) Energy = Power * Time Energy = 98.0 W * 24 hr * 3600 s/hr = 8.47 x 106 J Energy = 8.47 x 106 J / (4.186 x 103 J/kcal) = 2023 kcal Note: Calories given on food are actually kcal’s.

  16. CAPA # 10: Power Water flows over a waterfall which is 60 m high at an average rate of 8.0 x105 kg/s. If all of the potential energy of the water were converted into electric energy, how much electrical power could be produced by these falls? Power = Energy / Time Power = (mgh)/ Time = gh * m/t Power= 9.8 m/s2 * 60 m ( 8.0x105 kg/s) Power= 470x106 J/s = 470 MW

  17. Next Time • Chapter 6 – Work and Energy. • Quiz on Chapter 6 • Please see me with any questions or comments. See you Wednesday.

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