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Physics 151: Lecture 14 Today’s Agenda

Physics 151: Lecture 14 Today’s Agenda. Today’s Topics : One more problem with springs ! Power. Text Ch. 7.5 Energy and cars Text Ch. 7.6. Work & Power:. Two cars go up a hill, a BMW Z3 and me in my old Mazda GLC . Both have the same mass.

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Physics 151: Lecture 14 Today’s Agenda

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  1. Physics 151: Lecture 14Today’s Agenda • Today’s Topics : • One more problem with springs ! • Power. Text Ch. 7.5 • Energy and cars Text Ch. 7.6

  2. Work & Power: • Two cars go up a hill, a BMW Z3 and me in my old Mazda GLC. Both have the same mass. • Assuming identical friction, both engines do the same amount of work to get up the hill. • Are the cars essentially the same ? • NO. The Z3 gets up the hill quicker • It has a more powerful engine.

  3. Work & Power: • Power is the rate at which work is done. • Average Power is, • Instantaneous Power is,

  4. Work & Power: Units (SI) are Watts (W): Instantaneous Power: Average Power: • Power is the rate at which work is done. 1 W = 1 J / 1s Simple Example 1 : • A person of mass 80.0 kg walks up to 3rd floor (12.0m). If he/she climbs in 20.0 sec what is the average power used. W =  Fx dx P = W / Dt W = F h = (mg) h W = 80.0kg 9.8m/s2 12.0 m = 9408 J P = W / Dt = 9408 / 20.0s = 470 W

  5. ExampleProblem 7.41 • An energy-efficient light bulb, taking in 28.0 W of power, can produce the same level of brightness as a conventional bulb operating at power 100 W. The lifetime of the energy efficient bulb is 10 000 h and its purchase price is $17.0, whereas the conventional bulb has lifetime 750 h and costs $0.420 per bulb. • Determine the total savings obtained by using one energy-efficient bulb over its lifetime, as opposed to using conventional bulbs over the same time period. Assume an energy cost of $0.080 0 per kilowatt-hour.

  6. Solution:Problem 7.41

  7. Work & Power: Simple Example 2 : Engine of a jet develops a trust of 15,000 N when plane is flying at 300 m/s. What is the horsepower of the engine ? P = F v P = (15,000 N) (300 m/s) = 4.5 x 106 W = (4.5 x 106 W) (1 hp / 746 W) ~ 6,000 hp !

  8. ExampleProblem 7.47 • While running, a person dissipates about 0.600 J of mechanical energy per step per kilogram of body mass. If a 60.0-kg runner dissipates a power of 70.0 W during a race, how fast is the person running? Assume a running step is 1.50 m long. Solution:

  9. Lecture 14,ACT 3Work & Power • Starting from rest, a car drives up a hill at constant acceleration and then suddenly stops at the top. The instantaneous power delivered by the engine during this drive looks like which of the following, • A) • B) • C) Power time Power Z3 time Power time

  10. ExampleProblem 7.55 • A single constant force F (20.0 N) acts on a particle of mass m=5.00 kg. The particle starts at rest at t = 0. What is the power delivered at t = 3.00 s? Solution:

  11. v Lecture 14,ACT 4Power for Circular Motion • I swing a sling shot over my head. The tension in the rope keeps the shot moving in a circle. How much power must be provided by me, through the rope tension, to keep the shot in circular motion ? Note that Rope Length = 1m Shot Mass = 1 kg Angular frequency = 2 rads/sec A) 16 J/s B) 8 J/s C) 4 J/s D) 0

  12. V = 30 m/s Car 5 5% 100 Work & Power:Example 3 : • What is the power required for a car (m=1000 kg) to climb a hill (5%) at v=30m/s assuming the coefficient of friction m = 0.03 ? Ptot = Phorizontal + Pvertical v=const. - > a = 0 Phorizontal = F v = mmg vhorizontal Phorizontal ~ 0.03 (1000kg) (10m/s2 ) (30m/s) ~ 10 kW Pvertical = F v = mg vvertical = (1000kg) (10m/s2 ) (30m/s)(5/100) Pvertical ~ 15 kW Ptot ~ 10 kW + 15 kW = 25 kW

  13. Recap of today’s lecture • Work done by a spring • Power • P = dW / dt

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