is What 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 28always the same number.
It is an abstract thing in that it does not tell us
the mechanism or the reasons for the various
formulas.
Physics for Scientists &
Engineers 1
Fall Semester 2012
Lecture 12
Energy, Kinetic Energy, Work
February 7, 2012
Physics for Scientists&Engineers 1
1
K car = 1 mv 2 = 1 (1310 kg)(24.6 m/s)2 = 4.0 10 5 J
2
2
Mass of the Earth is 61024 kg, and it orbits the Sun with a
speed of 30,000 m/s
2
Unit of kinetic energy: [ K ] = [ m ] [ v]2 = kg m 2 / s 2
K sun = 1 mv 2 = 1 (6.0 10 24 kg)(3.0 10 4 m/s)2 = 2.7 10 33 J
2
2
Baseball (mass 5 ounces = 0.142 kg) thrown at 80 mph (=
35.8 m/s) has kinetic energy
This energy unit has received its own
name, Joule (J), named after British
physicist James Joule (1818-1889)
Energy unit 1 J = 1 kg m 2 / s2
Useful conversion: 1 J = 1 N m
K baseball = 1 mv 2 = 1 (0.142 kg)(35.8 m/s)2 = 91 J
2
2
Electron (me = 9.110-31 kg) moving with a speed of
1.3106 m/s (= 0.4% of the speed of light):
K e = 1 mv 2 = 1 (9.1 10 31 kg)(1.3 10 6 m/s)2 = 7.7 10 19 J
2
2
3
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Other Energy Units
4
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.
2
Use: v2 = v0 2 g( y y0 ) = 2 g( y0 y ) ( v0 = 0 )
Now use definition of kinetic energy and obtain in this case:
1 eV = 1.60210-19 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 = 4.01015 J
Physics for Scientists&Engineers 1
Physics for Scientists&Engineers 1
Example: Falling Vase
Atomic and nuclear physics introduced the electron-Volt
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2
Car of mass 1,310 kg driving 55 mph (=24.6 m/s)
Energy contained in the motion of an object
Definition K = 1 mv 2
Physics for Scientists&Engineers 1
Physics for Scientists&Engineers 1
Kinetic Energy Examples
Kinetic Energy
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K = 1 mv2 = 1 m ( 2 g( y0 y)) = mg( y0 y)
2
2
Numbers:
K = (2.40 kg) (9.81 m/s2 ) (1.30 m 0) = 30.6 J
5
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6
1
Work
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
K = mg( y0 y )
Gravitational force, Fg = mg, accelerated vase during free
fall, and thus increased its kinetic energy
Kinetic energy is proportional to the magnitude of the
gravitational force
Here the kinetic energy gained by the vase is simply the
product of the magnitude of the gravitational force and the
displacement
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7
Work Done by a Constant Force
In the previous example, the vase started out
with zero kinetic energy and after falling it had
gained a kinetic energy of 30.6 J
The gravitational force accelerates the vase and
gives it kinetic energy
We account for this change in the kinetic energy of
an object caused by a force with the concept of
work, W
Formal Definition of Work:
Work is the energy transferred to or from an object due
to the action of a force. Positive work is a transfer of
energy to the object, and negative work is a transfer of
energy from the object.
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Work Done by a Constant Force
Suppose we let the vase in the previous example slide from
rest along an inclined plane that has an angle with respect
to the horizontal
In our snowboarding example we showed that
a = g sin
In this example we have
= 90 so
a = g cos
We can determine the kinetic energy of the vase as a
function of the displacement using kinematics
2
v2 = v0 + 2ar
taking v0 = 0
v2 = 2 g cos r
1
K = mv2 = mg r cos
2
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Work Done by a Constant Force
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10
Work Done by a Constant Force
W = Fr
Because Fg = mg
The kinetic energy transferred to the vase was the result
of positive work done by the
force of gravity so
W = F r cos , where is the angle
between F and r
K = mg r cos = Wg
Limiting cases
February 7, 2012
W = Fr cos
This equation for the work done by a
constant force acting over some
spatial displacement holds for all
constant force vectors and arbitrary
displacement vectors
W =0
= 0 : Fg and r are parallel and pointing downward
Wg = mgr
= 90 : Fg points downward but the vase cannot move downward
Wg = 0
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12
2
Work Done by Constant Force - Example
Work as a Scalar Product
Using a scalar product, we can write the work done
by a constant force as
W = F r
W = F r cos = Fx x + Fy y + Fz z
)
(
13
)(
)(
)(
)
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14
Equivalent formulations (solving for K or K0):
K = K0 + W
K0 = K W
K K K 0 = W
This is the work-kinetic energy theorem
1212
mvx mvx 0 = max ( x x0 ) = Fx x = W
2
2
The work-kinetic theorem is equivalent to Newtons
Second Law
15
Work Done by Gravity
Kinetic energy cannot be less than zero
The work that can be done by an object (negative
work) cannot have a bigger magnitude than the
initial kinetic energy
The work-kinetic energy theorem is valid for
constant forces and variable forces
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16
Work Done Lifting an Object
Application of an external force F in vertical
direction
Work-kinetic energy theorem now includes work
done by gravity and work done by external force
Work done by gravitational
force during upward motion
of a projectile
Wg = -mgh
K K0 = Wg + WF
(reduces energy kinetic on the
way up)
When we lift an object, it is initially at rest (K0 = 0)
and also after we lifted it to its final height (K = 0)
=> W = W
Work done by gravitational
force on the projectile on
the way down
Wg = mgh
F
g
Lifting:
WF = -Wg = mgh
Lowering: WF = -Wg = -mgh
(increases kinetic energy on the
way down)
Physics for Scientists&Engineers 1
)(
Work-Kinetic Energy Theorem
Relationship between kinetic energy of an object
and the work done on the object by a constant
force
February 7, 2012
)(
W = 4.79 N 4.25 m + 3.79 N 3.69 m + 2.09 N 2.45 m
W = 1.25 J
Work-Kinetic Energy Theorem
Physics for Scientists&Engineers 1
W = F r = Fx rx + Fy ry + Fz rz
W = F r = Fx x = Fx ( x x0 )
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)
The work is given by
In one dimension
Physics for Scientists&Engineers 1
)
object of mass 18.0 kg, causing a displacement of that object
by r = 4.25,3.69, 2.45 m. What is the total work done by
(
Wnet = Fnet r = Fi r = Fi r = Wi
i
i
i
February 7, 2012
(
A constant force, F = 4.79, 3.79,2.09 N acts on an
this force?
The net work done by a net force is equal to the
sum of the work done by the individual forces
(
PROBLEM
17
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18
3
Example: Weightlifting
Lifting with Pulleys
Question:
Ronny Weller won silver in Sidney in 2000. He lifted 257.5 kg in the
jerk competition. Assuming he lifted weight to height of 1.83 m
and held it there, what was the mechanical work done by him in this
process?
Answer:
The work is
W = mgh
= (257.5 kg) (9.81 m/s ) (1.83 m)
2
19
Work Done by Variable Force
However, displacements
are different:
r1 = 2r2
So we get for work done
by T1:
x
x0
The work done can be represented by the area under the
curve of Fx(x) vs. x
21
Work Done by a Variable Force - Example
xi
Corresponds to the area under the curve in our sketch
SIMPLIFY
4.0
Physics for Scientists&Engineers 1
22
The units are good.
Lets calculate the area under the
curve
triangle
rectangle
1
W = ( 3.0 N ) ( 4.0 m ) + ( 4.0 m ) ( 2.0 N )
2
W = 12 J + 4.0 J = 16 J
xf
0.0
February 7, 2012
DOUBLE-CHECK
W = F ( x ) dx
( 3.0 + 0.50 x ) dx = [ 3.0 x ]
20
Work Done by a Variable Force - Example
RESEARCH
The work done by a variable force is given by
4.0
Identical work with or without pulleys!
Physics for Scientists&Engineers 1
Work Done by Variable Force - Example
W = Fx ( x ') dx '
Physics for Scientists&Engineers 1
February 7, 2012
A particle of mass m is subjected to a force acting
in the x-direction, Fx = (3.0+0.50x) N.
PROBLEM
Find the work done as the particle moves from x = 0.0 to
x = 4.0 m.
THINK
This forces varies with x, so we need to integrate over
the displacement.
SKETCH
Suppose the force acting on an object is not constant
The work done by such a force is
W=
W2 = T2 r2 = T2 r2 = mgr2
= mgr2 = W2
Physics for Scientists&Engineers 1
February 7, 2012
T1 = 1 T2 = 1 mg
2
2
Work done by T2:
W1 = T1 r1 = ( 1 T2 )(2r2 )
2
= 4.62 kJ
Note:
The work in lowering is then
-4.62 kJ.
February 7, 2012
Force needed to pull mass m up with pulleys is half of weight
4.0
+ 0.25 x2 0.0
0.0
W = ( 3) ( 4.0 ) + ( 0.25 ) (16 ) = 16 J
ROUND
W = 16 J
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4
The Spring Force
Work-Kinetic Energy Theorem for Variable Force
Assume a variable force in the x-direction
A very common variable force is the spring force
We can treat a spring in terms of motion and
forces in one dimension
Fx ( x ) = ma
We can use the chain rule to get
dv dv dx
=
dt dx dt
Integrate over the displacement to get the work
Choose the x-direction
a=
x
x
x
x0
x0
x0
W = Fx ( x ') dx ' = ma dx ' = m
We will use the spring force extensively
throughout the semester
So lets get a strong foundation now
dv dx '
dx '
dx ' dt
The spring force occurs in many physical situations
We start with a spring that is neither stretched
nor compressed
Change the variable of integration and do the integral
x
W = m
x0
v
v
v '2
dv dx '
dx ' = mv ' dv ' = m = K K0 = K
dx ' dt
2 v0
v0
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Physics for Scientists&Engineers 1
Equilibrium position of the end of the spring, x0
25
February 7, 2012
Spring Force
Physics for Scientists&Engineers 1
26
Spring
Spring Force
We can summarize by stating
Equilibrium position x = x0
The magnitude of the spring force is proportional to the
absolute value of the displacement from the equilibrium
position
The spring force is always in the opposite direction from
the displacement
Stretched to x = x1
Stretched to x = x2
Fs = k ( x x0 )
k is the spring constant
Always positive by definition
Has units of N/m
Compressed
The negative sign means that the spring force is always in
the opposite direction from the displacement
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Spring Force
Fs = kx
Hookes Law
Named after British physicist Robert Hooke (1635
1703), contemporary of Newton
The spring force always points toward the equilibrium
position
Restoring force
Linear restoring forces are found in many systems in nature
Crystal lattice
Shape deformations in nuclei
.
Physics for Scientists&Engineers 1
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Spring Constant
If we take x0 = 0, we have
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29
A spring has a length of
15.4 cm and is hanging
vertically from a support
point above it
A weight with a mass of
0.200 kg is attached to
the spring, causing it to
extend to a length of
28.6 cm
What is the value of the
spring constant?
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5
Spring Constant
Work Done by the Spring Force
We place the origin of our coordinate system so that
x0 = -15.4 cm and x = -28.6 cm
The spring force is
The work done by the spring in moving from x0 to x is
x
Physics for Scientists&Engineers 1
x0
x
1
12
WS = k x ' dx ' = kx2 kx0
2
2
x0
)
Taking x0 = 0 we get
( 0.200 kg ) 9.81 m/s2
Fs
mg
=
=
= 14.9 N/m
x x0
x x0
0.286 m ) ( 0.154 m )
(
February 7, 2012
x0
The work done by the spring is then
The force on the spring is provided by the weight of the
0.200 kg mass
We can solve for k
(
x
x0
Fs = k ( x x0 )
k=
x
WS = Fs ( x ') dx ' = ( kx ') dx ' = k x ' dx '
1
WS = kx2
2
31
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6