Transcript Physics 131: Lecture 14 Notes

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
Physics 151: Lecture 14, Pg 1
Work & Power:
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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.
Physics 151: Lecture 14, Pg 2
Work & Power:
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Power is the rate at which work is done.
Average Power is,
W
P
t
Instantaneous Power is,
dW
P
dt
Physics 151: Lecture 14, Pg 3
Work & Power:

Power is the rate at which work is done.
Average
Power:
W
P
t

Instantaneous
Power:
Units (SI) are
Watts (W):
dW
1 W = 1 J / 1s
P
dt
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 / t
W = F h = (mg) h
W = 80.0kg 9.8m/s2 12.0 m = 9408 J
P = W / t = 9408 / 20.0s = 470 W
Physics 151: Lecture 14, Pg 4
Example
Problem 7.41
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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.
Physics 151: Lecture 14, Pg 5
Solution:
Problem 7.41
P7.41 energy power time
For the 28.0 W bulb:
Energy used  28.0 W1.00  104 h  280 kilowatt  hrs
total cost  $17.00  280 kWh $0.080 kWh  $39.40
For the 100 W bulb:
Energy used  100 W1.00 104 h  1.00 103 kilowatt  hrs
1.00 104 h
# bulb us ed 
 13.3
750 h bulb

total cost

 13.3$0.420  1.00  103 kWh $0.080 kWh  $85.60
Savings with energy-efficient bulb  $85.60  $39.40  $46.20
Physics 151: Lecture 14, Pg 6
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 ?
dW
P
dt


W  F  x
 
dW  F  dx

dW  dx
P
F
dt
dt
 
P  F v
P=Fv
P = (15,000 N) (300 m/s) = 4.5 x 106 W
= (4.5 x 106 W) (1 hp / 746 W) ~ 6,000 hp !
Physics 151: Lecture 14, Pg 7
Example
Problem 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:
 1 step 
  24.0 J m
1.50
m


Concentration of Energy output  0.600 J kg  step 60.0 kg 
F  24.0 J m 1 N  m J  24.0 N
P  Fv
70.0 W  24.0 N v
v  2.92 m s
Physics 151: Lecture 14, Pg 8
Lecture 14, ACT 3
Work & Power
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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,
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A)
time
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B)
time
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C)
time
Physics 151: Lecture 14, Pg 9
Example
Problem 7.55
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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:
(a)
(b)
 F   F2 
P  Fv  Fv i  at  F0  t   t
 m   m 
20.0 N 2 
3.00 s   240 W
P  

 5.00 kg 

Physics 151: Lecture 14, Pg 10
Lecture 14, ACT 4
Power for Circular Motion
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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
v
Angular frequency = 2 rads/sec
A) 16 J/s
B) 8 J/s
C) 4 J/s
D) 0
Physics 151: Lecture 14, Pg 11
Work & Power:
Example 3 :
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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
5%
100
5
v=const. - > a = 0
Phorizontal = F v = m mg 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
Physics 151: Lecture 14, Pg 12
Recap of today’s lecture
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Work done by a spring
Power
P = dW / dt
Physics 151: Lecture 14, Pg 13