Presentation on theme: "Physics 151: Lecture 14, Pg 1 Physics 151: Lecture 14 Today’s Agenda l Today’s Topics : çOne more problem with springs ! çPower.Text Ch. 7.5 çEnergy and."— Presentation transcript:
Physics 151: Lecture 14, Pg 1 Physics 151: Lecture 14 Today’s Agenda l Today’s Topics : çOne more problem with springs ! çPower.Text Ch. 7.5 çEnergy and carsText Ch. 7.6
Physics 151: Lecture 14, Pg 2 Work & Power: l Two cars go up a hill, a BMW Z3 and me in my old Mazda GLC. Both have the same mass. l Assuming identical friction, both engines do the same amount of work to get up the hill. l Are the cars essentially the same ? l NO. The Z3 gets up the hill quicker l It has a more powerful engine.
Physics 151: Lecture 14, Pg 3 Work & Power: l Power is the rate at which work is done. l Average Power is, l Instantaneous Power is,
Physics 151: Lecture 14, Pg 4 Work & Power: l Power is the rate at which work is done. Instantaneous Power: Average Power: l 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 = F h = (mg) h W = 80.0kg 9.8m/s m = 9408 J W = F x dx P = W / t P = W / t = 9408 / 20.0s = 470 W Simple Example 1 : Units (SI) are Watts (W): 1 W = 1 J / 1s
Physics 151: Lecture 14, Pg 5 Example Problem 7.41 l 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 h and its purchase price is $17.0, whereas the conventional bulb has lifetime 750 h and costs $0.420 per bulb. çDetermine 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 $ per kilowatt-hour.
Physics 151: Lecture 14, Pg 6 Solution: Problem 7.41
Physics 151: Lecture 14, Pg 7 Work & Power: 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 ? Simple Example 2 : P = F v P = (15,000 N) (300 m/s) = 4.5 x 10 6 W = (4.5 x 10 6 W) (1 hp / 746 W) ~ 6,000 hp !
Physics 151: Lecture 14, Pg 8 Example Problem 7.47 l While running, a person dissipates about 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:
Physics 151: Lecture 14, Pg 9 Lecture 14, ACT 3 Work & Power l 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, l A) l B) l C) Z3 time Power time
Physics 151: Lecture 14, Pg 10 Example Problem 7.55 l 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:
Physics 151: Lecture 14, Pg 11 Lecture 14, ACT 4 Power for Circular Motion l 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 v A) 16 J/sB) 8 J/sC) 4 J/sD) 0
Physics 151: Lecture 14, Pg 12 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 = 0.03 ? Car 5% V = 30 m/s P tot = P horizontal + P vertical v=const. - > a = 0 P horizontal = F v = mg v horizontal P vertical = F v = mg v vertical = (1000kg) (10m/s 2 ) (30m/s)(5/100) P vertical ~ 15 kW P horizontal ~ 0.03 (1000kg) (10m/s 2 ) (30m/s) ~ 10 kW P tot ~ 10 kW + 15 kW = 25 kW
Physics 151: Lecture 14, Pg 13 Recap of today’s lecture l Work done by a spring l Power çP = dW / dt