Conservation of energy, power

1
Physics 100
Lecture 3
Conservation of
Mechanical Energy
January 30, 2017
2
Lifting an Object



Gravitational Potential
Energy (PE)
Removing the constraint
Energy conversion
3
Mechanical Energy

ME = KE + PE
• conservation principle
• bowling ball pendulum
•
•
(applet)
Bucket analogy
roller coaster & video
4
What speed will a frictionless roller coaster
have at 3.0 m altitude if it began its run from
rest at an altitude of 10 m?
100%
A.
B.
C.
D.
8.37 m/s
11.8 m/s
70.0 m/s
140 m/s
0%
A.
0%
B.
C.
0%
D.
5
What speed will a frictionless roller coaster
have at 3.0 m altitude if it began its run from
rest at an altitude of 10 m?
A.
B.
C.
D.
8.37 m/s
11.8 m/s
70.0 m/s
140 m/s
mghA  12 mvA2  mghB  12 mvB2
mghA  0  mghB  12 mvB2
ghA  0  ghB  12 vB2
2 g  hA  hB   vB2
2 g  hA  hB   vB  2 10 m/s 2  10  3.0 m 
vB  11.8 m/s
6
Conservation of Mechanical Energy


projectiles (applet)
mass on a spring (applet)
7
If you throw a ball upward with a speed of
17 m/s, how high does it go? (Try to use
conservation of energy to answer the question)
A.
B.
C.
D.
69%
0.85 m
1.70 m
14.5 m
28.9 m
10%
A.
14%
7%
B.
C.
D.
8
If you throw a ball upward with a speed of
17 m/s, how high does it go? (Try to use
conservation of energy to answer the question)
A.
B.
C.
D.
0.85 m
1.70 m
14.5 m
28.9 m
mghA  12 mvA2  mghB  12 mvB2
0  mv  mghB  0
1
2
2
A
v
17 m/s 
 hB 
2
2g
2 10 m/s 
2
A
hB  14.5 m
2
9
Review




Energy of motion = KE = ½ mv2
Energy of altitude = PE = mgh
Sum KE + PE = Mechanical energy
Mechanical energy is conserved if there
is no friction or other external force
besides gravity
10
Suppose a pendulum starts from rest at a height
of 0.30 m above its lowest point. What is its
speed when it swings through its lowest point?
A.
B.
C.
D.
2.45 m/s
3.00 m/s
7.74 m/s
10.0 m/s
83%
13%
3%
0%
A.
B.
C.
D.
11
Suppose a pendulum starts from rest at a height
of 0.30 m above its lowest point. What is its
speed when it swings through its lowest point?
A.
B.
C.
D.
2.45 m/s
3.00 m/s
7.74 m/s
10.0 m/s
mghA  12 mvA2  mghB  12 mvB2
mghA  0  0 
2 10 m/s
1
2
mvB2
2 ghA  vB
2
  0.30 m   v
B
 2.45 m/s
12
Power


The rate at which energy is transferred is
the power.
Because work is an energy transfer,
power is also the rate at which work is
done
work energy transfer Joule
Power 


 Watt
time
time
sec
13
Units of Power



The metric unit of power is the Watt (W)
The English unit of power is the
horsepower (hp)
For thermal energy we sometimes see
BTU/hr
14
Unit conversions
1 hp  746 W
BTU 1055 J 1 hr
1


 0.293 W
hr
BTU 3600 s
746 W 1 BTU/hr
1 hp 

 2545 BTU/hr
hp
0.293 W
15
What average power does a
weightlifter produce in order
to lift 1500 N (337 lb) a
distance of 1.5 m in
0.50 seconds?
A.
B.
C.
D.
97%
2250 W
4500 W
500 W
45 kW
3%
A.
0%
B.
C.
0%
D.
16
What average power does a
weightlifter produce in order
to lift 1500 N (337 lb) a
distance of 1.5 m in
0.50 seconds?
A.
B.
C.
D.
2250 W
4500 W
500 W
45 kW
W F d
P 
t
t
1500 N 1.5 m 

 4500 W
0.50 s
17
Another energy unit

Electricity is sold in units of kWh
W
P
t
 W  Pt  kW  h
1000 W 1 J/s 3600 s
1 kWh 


 3.60  106 J
1 kW
W
h
18
How many kWh of energy are in
a 287 (food) Calorie doughnut?
A.
B.
C.
D.
0.334 kWh
2.85 kWh
300 kWh
3.60×106 kWh
0%
A.
0%
B.
0%
C.
0%
D.
19
How many kWh of energy are in
a 250 (food) Calorie doughnut?
A.
B.
C.
D.
0.334 kWh
2.50 kWh
105 kWh
3.60×106 kWh
4186 J 1 kWh
287 Cal 

 0.334 kWh
6
Cal 3.6 10 J

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