physics 151: lecture 14, pg 1 physics 151: lecture 14 today’s agenda l today’s topics : çone...

Physics 151: Lecture 14, Pg 1

Physics 151: Lecture 14Physics 151: Lecture 14Today’s AgendaToday’s Agenda

Today’s Topics :One more problem with springs !Power. Text Ch. 7.5Energy and cars Text Ch. 7.6


Work & Power: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.



Power is the rate at which work is done. Average Power is,

Instantaneous Power is,tW

P

dtdW

P



Power is the rate at which work is done.

tW

P

dtdW

P

InstantaneousPower:

AveragePower:

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/s2 12.0 m = 9408 J

W = Fx dxP = W / t

P = W / t = 9408 / 20.0s = 470 W

Simple Example 1 :Simple Example 1 :

Units (SI) areWatts (W):

1 W = 1 J / 1s


ExampleExampleProblem 7.41Problem 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.


P7.41

energypowertime

For the 28.0 W bulb:

Energy used

28.0 W 1.00104 h 280 kilowatthrs

total cost

$17.00 280 kWh $0.080kWh $39.40

For the 100 W bulb:

Energy used

100 W 1.00104 h 1.00103 kilowatthrs

# bulb used

1.00104 h750 h bulb

13.3

total cost

13.3 $0.420 1.00103 kWh $0.080kWh $85.60

Savings with energy-efficient bulb

$85.60 $39.40 $46.20

Solution:Solution:Problem 7.41Problem 7.41



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 :Simple Example 2 :

dtdW

P xdFdWxFW

vFP

dtxd

FdtdW

P

P = F vP = (15,000 N) (300 m/s) = 4.5 x 106 W = (4.5 x 106 W) (1 hp / 746 W) ~ 6,000 hp !



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.

Concentration of Energy output

0.600 J kgstep 60.0 kg 1 step

1.50 m

24.0 J m

F 24.0 J m 1 Nm J 24.0 N

P Fv

70.0 W 24.0 N vv 2.92 m s

Solution:Solution:


Lecture 14,Lecture 14, ACT 3ACT 3Work & PowerWork & 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)

Z3

timePow

erP

owe r

Pow

e r

time

time



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?

(a)

P FvF vi at F 0Fm

t

F2

m

t

(b)

P 20.0 N 25.00 kg

3.00 s 240 W

Solution:Solution:


Lecture 14,Lecture 14, ACT 4ACT 4Power for Circular MotionPower 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 v

A) 16 J/s B) 8 J/s C) 4 J/s D) 0


Work & Power:Work & Power: Example 3 :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 ?

100

5Car5%

V = 30 m/s Ptot = Phorizontal + Pvertical

v=const. - > a = 0

Phorizontal = F v = mg vhorizontal

Pvertical = F v = mg vvertical = (1000kg) (10m/s2 ) (30m/s)(5/100)Pvertical ~ 15 kW

Phorizontal ~ 0.03 (1000kg) (10m/s2 ) (30m/s) ~ 10 kW

Ptot ~ 10 kW + 15 kW = 25 kW


Recap of today’s lectureRecap of today’s lecture

Work done by a spring Power

P = dW / dt

physics 151: lecture 14, pg 1 physics 151: lecture 14 today’s agenda l today’s topics : çone...

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