Download KE = 1 2 mv W = Fdx / W = F ⋅d x ∫

PHYS1110H, 2011 Fall.
Shijie Zhong
3. Work and Energy
Different types of energy:
a) Kinetic energy KE: energy of motion
b) Potential energy PE: stored energy, for example, gravitational PE,
electrostatic PE, elastic PE.
c) radiant energy: energy of light.
d) mass energy.
Energy can be converted from one form to another, but cannot be destroyed.
Kinetic energy of a mass m at a speed v:
1
KE = mv 2 .
(1)
2
The unit for energy is kg m2s-2 or Joule. 1 Joule = 1 kg m2s-2.
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Work done by a force F moving an object over a distance d in 1-D.
If the force F is constant and points at the same direction as the motion, then
the work is W = Fd .
If the force F is variable from initial position x1 to final position x2=x1+d,
and again assume that force F has the same direction as the motion, then we
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may break the path into n small segments, each with distance of Δx=d/n. The
force for i-th segment is Fi. The total work is
n
W = ∑ Fi Δx ,
(2)
i=1
or if we let Δx ->0, (2) can be turned into an integral expression:
x2
W = ∫ Fdx .
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(3)
x1
A more general expression for work done by force that is either along or
opposite to the direction of the motion is:
x2 

W = ∫ F ⋅ dx ,
(4)
x1
Note
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where a vector dot product is used. The work given by (4) can be negative, if
the force is at the opposite direction to the motion. Equation (4) can be
generalized to 2-D and 3-D.
Example 1. For a block pushed by a force F at a constant velocity for a
distance of d on a frictional surface that provides a frictional force f, the
work done by force F is positive and is W=Fd, but the work done by the
frictional force f is negative, Wf=-fd, because frictional force is at an
opposite direction to the motion. Since the block is at a constant velocity, its
acceleration is zero, and F=f in terms of magnitude, and W=-Wf.
Example 2. A block gets an initial push on a ramp with an angle θ from the
horizontal plane, and starts to move up the ramp for a distance d. What’s the
work done by the gravitational force?
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end 

W = ∫ F ⋅ dr ,
start


where dr is a vector along the ramp in the direction of motion, and F is the
gravitational force pointing down vertically.
 
π
F ⋅ dr = mgdr cos( + θ ) = −mgdr sin θ .
2
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Note
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end 
 d
W = ∫ F ⋅ dr = ∫ −mg sin θdr = −mgd sin θ .
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start
0
start
start
Alternatively, we can use vector notation for computing dot product and get
the
 same answer. Considering

F = −mgeˆ y , and dr = dxeˆ x + dyeˆ y ,
y 0 +d sin θ
end 
 end
W = ∫ F ⋅ dr = ∫ (−mgeˆ y ) ⋅ (dxeˆ x + dyeˆ y ) = −mg ∫ dy = −mgd sin θ .
y0
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Work-energy theorem in 1-D.
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Newton’s 2nd law:
dv
Fnet = ma = m .
dt
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Suppose that the mass m moves from positions x1 to x2 with net force Fnet
acting on it. Integrate (5) with respect to x from x1 to x2, and we have
x2
x 2 dv
(6)
∫ Fnet dx = ∫ m dx .
dt
x1
x1
dx
dt = vdt ,
(7)
dt
Substituting (7) back to (6) leads to
x2
t 2 dv
t 2 d(v 2 /2)
t
1
1
1
dt = mv 2 2 = mv2 2 − mv1 2 . (8)
∫ Fnet dx = m ∫ vdt = m ∫
t1 2
dt
2
2
x1
t1
€ t1 dt
x2
1
2 1
2
(9)
∫ Fnet dx = mv2 − mv1 .
2
2
x1
We can interpret (9) as the work done by net force from x1 to x2 is the
change of kinetic energy from x1 to x2 or
We always have:
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(5)
dx =
W1−2 = KE 2 − KE1
This is the work-energy theorem.
Some examples and applications:
Example 3. A ball with mass m is thrown up vertically at an initial speed v0.
The gravitational force is mg and points down. What’s the speed after the
ball travels up for a distance of d?
Note
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Using the work-energy theorem (10), in a coordinate system where positive
y points upward, the work done by the net force, i.e., gravitational force mg,
from the initial position to the final position of distance d, W=-mgd, which
must be equal to the change of kinetic energy between these two positions.
1
1
−mgd = mv f 2 − mv0 2 ,
(11)
2
2
(12)
v f 2 = v0 2 − 2gd .
The expression (12) is the same as that from the kinematics.
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Example 4. Spring-block system. When a spring is compressed by a
distance x with an external force Fext, the spring pushes back with a force
Fspring. Hooke’s law says that the spring force,
Fspring = −kx ,
(13)
where k is the spring constant. The origin of the coordinate (i.e., x=0) is
where the spring is free with no force (i.e., neither compressed or extended).
The negative sign in equation (13) is important in that it defines the direction
of the spring force. That is, when the spring is compressed, x is negative,
and spring force is positive according to (13), i.e., pointing to positive x
direction. Similar argument can be made for when the spring is extended.
Now a block is attached to a spring and is placed on a frictionless surface. A
external force is applied to the block to compress the spring by a distance of
h. Then the external force (e.g., your hand) is removed, and the spring
pushes the block back to the neutral or equilibrium position of the spring
(i.e., at x=0). What is the velocity of the block at the neutral position?
Note
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Let’s first compute the work done by the spring force to the block from x=-h
to x=0.
0
0
0 1 2
1
W = ∫ Fspring dx = ∫ −kxdx = − kx 2
= kh . (14)
−h 2
2
−h
−h
According to the work-energy theorem,
1 2
1
kh = KE x=0 − KE x=−h = mv f 2 ,
(15)
2
2
where we considered the zero kinetic energy at x=-h.
Therefore,
k
vf =
h.
(16)
m
Example 5. Escape velocity in a gravitational field. For an object with mass
m in the gravitational field of a planet with mass M and radius Rp, the
gravitational force acting on the object is given by the Newton’s law of

GmM 
gravitation as F = − 2 er , where r is the distance between the object and
r

the center of the planet, and er is the unit vector pointing from the center of
the planet to the object (i.e., the force points to the center of the planet).
What is the escape velocity for the object? What’s the energy needed to let
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the object escape the planet’s gravitational field?
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Suppose that vi is the velocity of the object at the surface of the planet (i.e.,
launching velocity at r= Rp), and vf is the velocity at some distance away
from the planet (i.e., r=Rf). In the traveling process, the gravitational force
does a negative work to the object. The work-energy theorem states:
  1
1
2
2
∫ F ⋅ dr = mv f − mvi .
2
2
Rp
Rf
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Note
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(17)
5
Rf
∫ −
Rp
GMm
1
1
2
dr
=
mv
−
mvi 2 .
f
2
2
2
r
Rf
1
1
1
1 2 1 2
dr
=
GM
(
−
)
=
v f − vi .
2
R f Rp
2
2
Rp r
1
1
vi 2 = 2GM (
−
) + vf 2
(20)
Rp R f
.
−GM ∫
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(18)
(19)
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Equation (20) applies to any distance that the object travels after the launch.
For escape velocity, we may set vf=0, and Rf to infinite.
2GM
vi 2 =
(21)
Rp .
We may write (21) as
(22)
vi 2 = 2gR p ,
where we considered that gravitational acceleration g is
GM
g= 2 ,
(23)
Rp
For the earth, g=9.8 m/s2, and Rp=6.4x106 m, using (22),
we have vi=11 km/s.
The energy needed is 1/2mvi2. For a 50 kg object, the energy is 3.0x109 J.
Generalize to 3-D
Work done in 3-D
For a force moving an object from positions 1 to 2 along a 3-D trajectory,
the work done is
Note
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
r2
 
W = ∫ F ⋅ dr ,
(24)
r1


where r1 and r2 are position vectors for positions 1 and 2 and integral is
along a given path or trajectory (e.g., as in the left figure).
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Even if the starting and ending positions are the same, the integral may not
be the same if a different path or trajectory is taken, for example, for the
solid curve in the right figure. Therefore, sometimes we use the following
expression
 to show that the integral is along a path C,
W = ∫ F ⋅ dr ,
(25)
C
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Work-energy theorem in 3-D
Newton’s 2nd law in 3-D:


dv
F=m ,
(26)
dt


 

r2 
r2 dv
m t 2 dv 2
1
1
 r2 dv 

⋅ v dt = ∫
dt = mv2 2 − mv1 2 ,
∫ F ⋅ dr = ∫ m ⋅ dr = m ∫
dt
2 t1 dt
2
2
r1
r1
r1 dt
1
1
W1−2 = mv2 2 − mv1 2 ,
2
2
(28)
W1−2 = KE 2 − KE1 .
(27)
Equation (28) is the work-energy theorem in 3-D.
Path-dependent and path-independent integrals
Example 6. Work done by gravitational force to move an object from
positions 1 to 2 along different paths.
Note
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
In example 2, we expressed dr = dxeˆ x + dyeˆ y , and for this problem because
the gravitational force always points to the center (i.e., center of planet
where we use as the origin of the coordinates), it is more convenient to
express

dr = dreˆr + rdθeˆθ ,€ (29)
where eˆr ,and eˆθ are unit vectors in polar coordinates, and are
perpendicular to each other.
 
GMm
W = ∫ F ⋅ dr = ∫ (− 2 eˆr ) ⋅ (dreˆr + rdθeˆθ ) ,
r
C
C
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r2
GMm
W = ∫ (− 2 )dr ,
(30)
r
r1
where we used eˆr ⋅ eˆθ = 0 , and eˆr ⋅ eˆr = 1. From (30), we have
1 1
(31)
W = GMm( − ) .
r2 r1
Since€in deriving (31),
€ we did not need to consider the specific path from
points 1 to 2, we may conclude that the work done by gravitational force
only depends on the radial locations r of the starting and ending points, i.e.,
r1 and r2, and DOES NOT depend on the path or how the object gets from
points 1 to 2.
Remarks: This path-independent work done by a force is controlled by the
property of the force. This type of force is called conservative force.
Gravitational force and electrostatic force are conservative forces.
Not all the forces’ work is path-independent, and NOT all the forces are
conservative forces. For example, the work done by frictional force clearly
depends on the path. The work done by frictional force to your car when you
drive from Boulder to Denver using US36 is clearly different from (i.e.,
Note
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smaller than) that if you drive your car from Boulder to Fort Collins and
then to Denver.

Example 7. Consider work done by a force F = xyeˆ x + y 2 eˆ y , to move a
particle from points (0,0) to (0,1)
 along two different paths, paths 1 and 2.
Then consider another force F = Aeˆ y (i.e., constant force) for the same
paths.
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For path 1,
 
 
 
 
W = ∫ F ⋅ dr = ∫ F ⋅ dr + ∫ F ⋅ dr + ∫ F ⋅ dr .
path1
a
b
c



Along line a, dr = dxeˆ x , line b, dr = dyeˆ y , and line c, dr = −dxeˆ x .
1
1
  1
1
W = ∫ F ⋅ dr = ∫ x ⋅ 0dx + ∫ y 2 dy + ∫ −x ⋅1dx = − .
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path1
0
0
0
For €
path 2,
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  1 2

1

W = ∫ F ⋅ dr = ∫ y dy = ≠ ∫ F ⋅ dr .
3 path1
path2
0
Therefore,
this force is non-conservative.

For F = Aeˆ y ,
  1
W = ∫ F ⋅ dr = ∫ Ady = A .
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path1
€
 
∫ F ⋅ dr = ∫ Ady = A =
W=
path2
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0
1
0
 
∫ F ⋅ dr .
path1
Therefore, constant force is a conservative force.
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Note
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