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February 8, 2012
Physics for Scientists&Engineers 1
1
Physics for Scientists &
Engineers 1
Spring Semester 2012
Lecture 15
Power and Potential Energy
February 8, 2012
Physics for Scientists&Engineers 1
2
Power for a Constant Force
Work for a constant force:
Differential:
(because
F
=constant)
According to definition of power:
Final answer:
Or:
W
=
F
• Δ
r
dW
=
F
•
d
r
P
=
dW
dt
=
F
•
d
r
dt
P
=
F
•
v
P
=
Fv
cos
α
Fv
February 8, 2012
Physics for Scientists&Engineers 1
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Accelerating Car
Question:
Car of mass 3,120 lb (= 1,415 kg) can reach the 60 mph (= 26.8 m/s)
mark in 4.0 seconds
What is the average power needed to accomplish this?
Answer:
Car’s kinetic energy at 60 mph
Workkinetic energy theorem:
K
=
1
2
mv
2
=
1
2
(1415 kg)
⋅
(26.8 m/s)
2
=
508 kJ
W
= Δ
K
=
K
−
K
0
=
508 kJ
February 8, 2012
Physics for Scientists&Engineers 1
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Accelerating Car
Average power to get to the 60 mph mark in 4.0 seconds:
Discussion:
Is this realistic?
•
No! 170 hp are not enough to accomplish this acceleration for a car
of mass 1415 kg
• This Corvette has 430 hp
•
Why?
Not perfectly efficient
engine; not able to produce
peak hp at all rpm; energy
loss from friction and drag
•
Done
P
=
W
Δ
t
=
5.08
⋅
10
5
J
4.0 s
=
127 kW
=
170 hp
February 8, 2012
Physics for Scientists&Engineers 1
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Power to Keep a Car Moving
A car travels at 60.0 mph on a level road.
The car has a drag
coefficient of 0.33 and a frontal area of 2.2 m
2
.
How much power
does the car need to maintain its speed?
Take the density of air to
be 1.29 kg/m
3
.
We can write the power as
The force here is the force of air resistance
We can combine these equations to get
So the power required to overcome air resistance is
P
=
Fv
=
F
drag
v
F
drag
=
Kv
2
K
=
1
2
c
d
A
ρ
air
P
=
1
2
c
d
A
air
v
2
v
=
1
2
c
d
A
air
v
3
P
=
1
2
0.33
()
2.2 m
2
1.29 kg/m
3
26.8 m/s
3
=
9.01 kW
=
12.1 hp
Lifting Bricks
PROBLEM
•
A load of bricks at a construction site has a mass of 85.0 kg. A crane
raises this load from the ground to a height of 50.0 m in 60.0 s at a
low constant speed. What is the average power of the crane?
THINK
•
Raising the bricks at a low constant speed means that the kinetic
energy is negligible, so the work in this situation is done against
gravity only
•
There is no acceleration, and friction is negligible
•
The average power then is just the work done against gravity divided
by the time it takes to raise the load of bricks to the stated height
SKETCH
•
In the free body diagram, the tension on the cable
is equal to the weight of the bricks
•
The load is moved vertically a distance
h
2/8/2012
Physics for Scientists&Engineers 1
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Lifting Bricks
RESEARCH
•
The work done by the crane is
•
The average power required to lift the load in the given time is
SIMPLIFY
CALCULATE
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W
=
mgh
P
=
W
Δ
t
P
=
mgh
Δ
t
P
=
85.0 kg
()
9.81 m/s
2
50.0 m
60.0 s
=
694.875 W
Lifting Bricks
ROUND
DOUBLECHECK
•
Convert out result for the required average power from
watts to horsepower
•
So a 1 hp motor would be sufficient to lift the 85.0 kg
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This note was uploaded on 04/02/2012 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.
 Spring '08
 Wolf
 Physics, Force, Power

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