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Energy Conversions and Efficiency: Pendulums and Calculating Efficiency

This chapter explores the energy conversions and efficiency in simple systems such as pendulums. It provides examples and calculations to understand the concepts. It also discusses energy efficiency and its impact on the environment.

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Energy Conversions and Efficiency: Pendulums and Calculating Efficiency

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  1. Chapter 6 Energy Conversions and Efficiency

  2. Simple Systems: Pendulums • Pendulums continuously convert between Ep and Ek • When we release a pendulum, Ep is converted to Ek until it reaches the lowest point (2). • At this point the pendulum is at Vmax and all of its energy is Ek • At the highest points, the pendulum stops momentarily before changing direction (1 and 3) Here all of its energy is Ep

  3. Pendulum simulation If we ignore energy loss to friction, total energy is equal anywhere along the pendulum's path EP= mg∆h EP= mg∆h Ek = ½ mv2

  4. Example • A pendulum with a mass of 1.0 kg is raised 0.5m above its resting point. Calculate the maximum speed of the pendulum.

  5. At the top, all energy is Ep Ep = mg∆h = (1.0Kg)(9.81m/s2)(0.5m) = 4.905 J • Max speed is at the bottom where all energy is Ek and we now know how much energy is in the system. • Ek = ½ mv2 4.905 J = ½(1.0Kg)(v)2 V= 3.312m/s

  6. A 35 kg girl on a trampoline gains 500 J of gravitational potential energy at the top of her bounce. Find the elastic potential energy of the trampoline

  7. A 35 kg girl on a trampoline gains 500 J of gravitational potential energy at the top of her bounce. Find the elastic potential energy of the trampoline • 500 J

  8. A 35 kg girl on a trampoline gains 500 J of gravitational potential energy at the top of her bounce. How high does she bounce?

  9. A 35 kg girl on a trampoline gains 500 J of gravitational potential energy at the top of her bounce. How high does she bounce? Ep = mgh h = Ep = 500 J = 1.5 m m (35kg)(9.81m/s2)

  10. Video MonD Physics of Fun

  11. 6.1 Efficiency of Energy Conversions Calculating Efficiency • Efficiency is a ratio and is expressed as a percent. • Output and input energies must use the same units. or Efficiency = useful output energy × 100% total input energy Efficiency = Eout × 100% Ein

  12. A cyclist uses 52 kJ of energy in a 100 km race. If he averages 50 km/h and the total mass of bike and rider is 80 kg, what is his % efficiency?

  13. A cyclist uses 52 kJ of energy in a 100 km race. If he averages 50 km/h and the total mass of bike and rider is 80 kg, what is his % efficiency? • Ein = 52 kJ = 52 000 J • Eout = Ek • Vave = 50km/h = 13.88889 m/s • m = 80 kg

  14. A cyclist uses 52 kJ of energy in a 100 km race. If he averages 50 km/h and the total mass of bike and rider is 80 kg, what is his % efficiency? • Ein = 52 kJ = 52 000 J • Eout = Ek • Vave = 50km/h = 13.88889 m/s • m = 80 kg Ek = ½ mv2

  15. A cyclist uses 52 kJ of energy in a 100 km race. If he averages 50 km/h and the total mass of bike and rider is 80 kg, what is his % efficiency? • Ein = 52 kJ = 52 000 J • Eout = Ek • Vave = 50km/h = 13.88889 m/s • m = 80 kg Ek = ½ mv2 = ½ (80kg)(13.88889m/s)2 = 7716 J

  16. A cyclist uses 52 kJ of energy in a 100 km race. If he averages 50 km/h and the total mass of bike and rider is 80 kg, what is his % efficiency? • Ein = 52 kJ = 52 000 J • Eout = Ek • Vave = 50km/h = 13.88889 m/s • m = 80 kg Ek = ½ mv2 = ½ (80kg)(13.888889m/s)2 = 7716 J Efficiency = Eout Ein

  17. A cyclist uses 52 kJ of energy in a 100 km race. If he averages 50 km/h and the total mass of bike and rider is 80 kg, what is his % efficiency? • Ein = 52 kJ = 52 000 J • Eout = Ek • Vave = 50km/h = 13.88889 m/s • m = 80 kg Ek = ½ mv2 = ½ (80kg)(13.888889m/s)2 = 7716 J Efficiency = Eout = 7716 J x 100 % = 15% Ein 52 000 J

  18. Look at Model Problem 1 page 225 – 226 • Do Practice Problems page 227 # 1-9

  19. Do Workbook BLM 6-2 Calculating Efficiency • Do Check Your Understanding p. 231 #1 - 5, 7

  20. 6.2 Energy Efficiency and the Environment • The average car is only 20% efficient. • Incandescent light bulbs are only 5% efficient. • Compact fluorescents are 20% efficient • For every litre of gasoline burned 2.4 kg of CO2 is released into the atmosphere. • Read P. 232 – 241 and do P. 242 # 1-9

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