PHY183-Lecture14pre - What is Energy Richard Feynman talks...

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February 7, 2012 Physics for Scientists&Engineers 1 1
Physics for Scientists &
Engineers 1
Fall Semester 2012
Lecture 12
Energy, Kinetic Energy, Work
What is Energy?
 Richard Feynman talks about the concept of
energy:
• “It is important to realize that in physics today,
we have no knowledge of what energy is. We do
not have a picture that energy comes in little
blobs of a definite amount. It is not that way.
However, there are formulas for calculating
some numerical quantity, and when we add it all
together it gives ‘28’—always the same number.
It is an abstract thing in that it does not tell us
the mechanism or the reasons for the various
formulas.”
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Kinetic Energy
 Energy contained in the motion of an object
 Definition
 Unit of kinetic energy:
 This energy unit has received its own
name, Joule (J), named after British
physicist James Joule (1818-1889)
 Energy unit
 Useful conversion: 1 J = 1 N m
K = 1
2
mv 2
[ K ] = [ m ] ⋅ [ v ] 2 = kg m 2 /s 2
1 J = 1 kg m 2 2
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Kinetic Energy Examples
 Car of mass 1,310 kg driving 55 mph (=24.6 m/s)
 Mass of the Earth is 6·10 24 kg, and it orbits the Sun with a
speed of 30,000 m/s
 Baseball (mass “5 ounces” = 0.142 kg) thrown at 80 mph (=
35.8 m/s) has kinetic energy
 Electron (
m
e
= 9.1·10 -31 kg) moving with a speed of
1.3·10
6
m/s (= 0.4% of the speed of light):
K
car
= 1
2
mv 2 = 1
2
(1310 kg)(24.6 m/s) 2 = 4.0 ⋅ 10 5 J
K
sun
= 1
2
mv 2 = 1
2
(6.0 ⋅ 10 24 kg)(3.0 ⋅ 10 4 m/s) 2 = 2.7 ⋅ 10 33 J
K
baseball
=
1
2
mv 2 =
1
2
(0.142 kg)(35.8 m/s) 2 = 91 J
K
e
= 1
2
mv
2
= 1
2
(9.1 ⋅ 10
− 31
kg)(1.3 ⋅ 10
6
m/s)
2
= 7.7 ⋅ 10
− 19
J
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Other Energy Units
 Atomic and nuclear physics introduced the electron-Volt
1 eV = 1.602·10 -19 J
(electron in previous example has ~5 eV kinetic energy)
 The energy you eat is measured in food calories
1 Cal = 4186 J
(1 Cal > 40 times the kinetic energy of the baseball in
previous example)
 For very large energy measures one need a large unit. The
energy released by one million tons of TNT is large
1 Mt = 4.0·10 15 J
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Example: Falling Vase
Question:
 A crystal vase (mass 2.40 kg) is dropped from a height of 1.30 m
and falls to the floor. What is its kinetic energy just before impact?
Answer:
 Once we know the velocity of the vase just before impact, we can
put it into our equation for the kinetic energy. To obtain this
velocity, we remind ourselves of the kinematics of free-falling
objects.
 Use:
 Now use definition of kinetic energy and obtain in this case:
 Numbers:
v 2 = v
0
2 − 2 g ( y − y
0
) = 2 g ( y
0
− y ) v
0
= 0
()
K =
1
2
mv 2 =
1
2
m 2 g ( y
0
− y ) = mg ( y
0
− y )
K = (2.40 kg) ⋅ (9.81 m/s
2
) ⋅ (1.30 m − 0) = 30.6 J

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Example: Falling Vase - Observations
 In the case of the falling vase, the kinetic energy is a
function of the height (linearly dependent on height) from
which the vase was released
 Gravitational force, F
g
=mg, accelerated vase during free