Download 16 Chapter 7A Work-Energy Theorem.pages

Chapter 7A Work-Energy Theorem!
!
There are three parts to this. What is energy? What is work? And how are they related?!
Energy!
Energy is the ability of an object to affect the motion of another object. This is a new way of
looking at an object. Instead of objects applying forces on each other, energy is transferred
between object. The transfer is still based on forces but the detail can be unnecessary.!
The simplest type of energy is stored in the motion (velocity) of an object. A moving object has the
ability of move another object. This type of energy is called kinetic energy.!
Work!
There are several requirements to change the kinetic energy of an object. First, a force must be
applied. Force produces an acceleration which changes the velocity.!
force
K=0
force
K>0
velocity
!
However, a force by itself is not enough. The object must have moved.!
force
K=0
force
K=0
!
This means that the object must have moved.
object.!
force
There is a displacement that occurred by the
K=0
force
displacement
K>0
velocity
!
Finally, the displacement must have be due to the force. Here is a case where it isn’t.!
force
velocity, displacement
K
!
The force must be responsible for the displacement. The more the displacement is in the direction
of the force, the more the force is responsible for the displacement.
page 1
These requirements are put together into a model for how much energy is transferred.
amount of energy transferred by a force F is this.!
!
!
W = F Δx cos θbetween
!
!
F and Δx
The
!
This is the work done by the force F. A nice characteristic of energy is the fact that they just add.
The total work done by all of the forces on the object is!
Wtotal = ΣW !
!
!
Wtotal = ΣF Δx cos θbetween
!
!
ΣF and Δx
!
With this application of work, the kinetic energy of the object changes. Here is how the two are
related.!
Vector Dot Product!
The above total work equation can be written as a vector equation. It is describe precisely by the
dot product operation. The result of this operation is a scalar. The total work can be written as
this vector equation.!
!
!
Wtotal = ΣF ⋅ Δx !
Here is the vector definition of the dot product.!
! !
A ⋅ B = Ax Bx + Ay By !
Here is the trigonometric definition of the dot product.!
! !
! !
A ⋅ B = A B cos θbetween
!
!
A and B
!
A more meaningful way to look at the dot product is to see it an a projection. Let’s say we have
these two vectors.!
B
A
θbetween A and B
!
The three factors in the trigonometric definition above can be grouped into two factors. You can
group the magnitude of A and the cosine together or the magnitude of B and the cosine together.!
! !
!
A ⋅ B = A cos θbetween
!
!
A and B
!
!
⋅ B = B cos θbetween
!
!
A and B
!
⋅ A !
This first term of the first grouping is the component of A in the direction of B. This first term of the
second grouping is the component of B in the direction of A.!
B
A
|B|•cos(θbetween A and B)
B
A
|A|•cos(θbetween A and B)
!
Both are valid descriptions of the same thing.!
page 2
Work Done by a General Force!
To find the work done by any force or net force, we would says that the force is a general function
of the position and we would integrate it along the path of travel. Here, r is just a general position.
This is the formal, general definition of work.!
!
rf
W =
!
!
∫! F(x) ⋅ dr !
ri
!
rf
Wtotal =
!
!
∫! ΣF(x) ⋅ dr !
ri
Work-Energy Theorem!
Using the general definition above, the total work along any one direction (call it x) is!
!
rf
Wtotal =
∫!
!
!
ΣF(x) ⋅ dr =
ri
!
rf
∫!
! !
ma ⋅ dr =
ri
xf
∫ ma ⋅ dx !
xi
Using the definition of the acceleration, this becomes!
xf
Wtotal =
dv
∫ m dt ⋅ dx !
xi
Using the chain rule, we can write the velocity derivative in the following way. Rearranging terms
and integrating, we get this.!
xf
Wtotal =
∫
xi
dv dx
m
⋅ dx =
dx dt
xf
∫
xi
dx dv
m ⋅ dx =
dt dx
vf
∫
vi
1
mv ⋅ dv = mv 2
2
vf
vi
1
1
= mv f2 − mvi2 !
2
2
The total work produces a difference in the value of the term that looks like this.
depends on the speed of the object. We call this term the kinetic energy, K.!
This term
1
K = mv 2 !
2
The total work on an object produces a change in the kinetic energy of theta object.!
Wtotal = ΔK !
This statement is called the work-kinetic energy theorem. What you need are two states of the
object where you can measure the speed or the kinetic energy of the object.!
The one thing that this approach does not do is tell you anything about how much time is required
for a process to complete. For that you would have to go back to the equations of motion or
forward to a concept call momentum.!
Unit!
The unit of energy is the joule. We have two ways to form it, from work and from kinetic energy.!
⎡W ⎤ = ⎡ F ⎤ ⎡ x ⎤ = Nm ≡ J !
⎣ ⎦ ⎣ ⎦⎣ ⎦
2
⎡ K ⎤ = ⎡ m ⎤ ⎡ v ⎤ 2 = kg m = kg m m = Nm ≡ J ⎣ ⎦ ⎣ ⎦⎣ ⎦
s2
s2
page 3
Example: Ramp!
A 5.0 kg block on a 3.0 meters long 20° ramp slides down when released from rest.
coefficient of kinetic friction between the block and the ramp is 0.20.!
The
5.0 kg
3.0 m
20°
!
• What is the work done on the block by friction?!
Here is how to do this using the magnitude definition. The magnitude of the force is !
!
fk = µkmg cos θ !
The magnitude of the displacement is !
!
Δr = 3 !
The angle between the force and displacement is 180° so we have!
cos(180°) = −1 !
The work is!
Wf = (µkmg cos θ)(3)(−1) = −(0.2)(5)(9.8)cos 20°(3) = −27.6 J !
k
A negative work done on an object is a removal of energy from that object.!
!
Here is the formal way of doing this. Starting with the definition of work,!
Wf =
k
!
!
∫ fk ⋅ dr !
Let me set the direction down the ramp as the positive x axis. !
+x
dr
path
fk
!
The frictional force is!
!
fk = −µk N iˆ = −µkmg cos θ iˆ !
The path of the block is!
!
dr = dr iˆ where 0 ≤ r ≤ 3 !
The work done is!
3
Wf =
k
!
3
∫ −µkmg cos θ iˆ ⋅ dx iˆ = −µkmg cos θ ⋅ x 0 = −(0.2)(5)(9.8)cos 20°(3) = −27.6 J !
0
page 4
• What is the work done on the block by gravity?!
Let’s calculate the work using that magnitude definition.!
!
!
Wg = Fg Δr cos θbetween
!
!
Fg and Δr
!
Here is a diagram of the relevant parts.!
dr
W
path
θbetween Fg and ∆r
!
The magnitude of the force is !
!
Fg = mg !
The magnitude of the displacement is !
!
Δr = 3 !
The angle between the force and displacement is the angle complementary to 20° so we have!
cos(70°) = sin(20°) !
The work is!
Wg = (mg)(3)sin 20° = (5)(9.8)(3)sin 20° = 50.3 J !
A positive work done on an object is a addition of energy to that object.!
!
Here is the formal way of doing this. Starting with the definition of work,!
Wg =
!
!
∫ fk ⋅ dr !
Here is my coordinate system. !
+y
dr
+x
W
path
!
The force of gravity is!
!
Fg = mg sin θ iˆ − mg cos θ jˆ !
The path of the block is!
!
dr = dr iˆ where 0 ≤ r ≤ 3 !
The work done is!
3
Wg =
∫
0
3
ˆ ⋅ (dr iˆ) =
(mg sin θ iˆ − mg cos θ j)
3
∫ mg sin θ ⋅ dr = mg sin θ ⋅ r 0 = (5)(9.8)sin 20°(3) = 50.3 J 0
page 5
• What is the work done on the block by the normal force?!
Let’s start with a diagram.!
θbetween N and ∆r
N
dr
path
!
The magnitude of the force is!
!
N = mg cos θ !
The magnitude of the displacement is!
!
Δr = 3 !
The angle between the normal force and the displacement is 90°.!
cos(90°) = 0 !
The work is!
!
WN = (mg)cos 20°(3)(0) = 0 J !
• What is the total work done on the block?!
This is just the addition of our results. This is!
!
Wtotal = −27.6 J + 50.3 J + 0 J = 22.7 J !
• What is the final speed of the block? !
We would expect the block to be moving faster than it was since the work is positive. Its final
speed comes from the work-energy theorem.!
Wtotal = ΔK !
!
1
1
22.7 J = K f − Ki = K f = mv f2 = (5)v f2
2
2
⇒ v f = 3.01 m / s !
page 6
• Check this using Newton’s second law.!
Here is the free-body diagram.!
N
+y
fk
+x
W
path
!
The net force in the y direction is zero.!
ΣFy = N − mg cos θ = 0
⇒ N = mg cos θ !
The net force in the x direction is!
ΣFx = mg sin θ − µk N = mg sin θ − µkmg cos θ = mg(sin θ − µk cos θ) = max !
The acceleration is!
ax = g(sin θ − µk cos θ) = (9.8)(sin 20° − (0.2)cos 20°) = 1.51 m / s 2 !
The block travels 3 meter s from rest and since the acceleration is constant, we can do this.!
Δ(v 2 ) = 2aΔx
!
!
!
⇒ v f2 = 2(1.51)(3) ⇒ v f = 3.01 m / s !
page 7
Work Done by a Spring!
Here is a new type of force. The force exerted by a spring on an object. Here, he spring is NOT
an object. It is a source of force like the Earth for gravity and a rope for tension.!
The force a spring exerts is modeled by Hooke’s law. It says that the force from a spring is
proportional to the amount of compression or stretch of the spring from its equilibrium position.
The equilibrium position of a spring is the position at which the spring can remain stationary.!
equilibrium!
position
x=0
Fspring = 0
x<0
Fspring
x>0
Fspring
!
Hooke’s law says this. Here, the vector x is understood to be the displacement from equilibrium.!
!
!
Fspring = −kx !
The constant k is called the spring force constant. It tells you how stiff a spring is. The reason we
didn’t deal with this earlier is because the force, and thus the acceleration, is not constant when a
spring is involved. The above model produces sinusoidal motion, motion described by sine and
cosine functions.!
For an object subjected to only the spring force, the second law says this.!
!
!
!
!
ΣF = Fspring = −kx = ma
⇒ x"" = −
k
x!
m
The general solution to the above differential equation is this with two constants.!
x(t) = C1 sin(ωt + C 2 ) where ω =
k
!
m
We can just apply the definition of work to the spring force to figure out how energy is transferred.!
!
page 8
Example!
A spring with a spring force constant of 1000 N/m is compressed by 0.10 meter with a mass of
0.50 kg attached. !
• How much work is done by the spring?!
x<0
Fspring
x=0
!
The initial state of the mass is at rest so its initial kinetic energy is zero. The final state is at the
equilibrium position where the kinetic energy is unknown. The work done by the spring on the
mass is the following.!
Because the force is variable, we can’t say what the magnitude of the force is.
integrate the dot product. Let the right be the positive direction.!
We have to
!
!
Fspring = −kx = −kx iˆ !
!
dx = dx iˆ where − 0.10 ≤ x ≤ 0 !
The work done is!
0
Wspring =
∫
0
(−kx iˆ) ⋅ (dx iˆ) =
−0.10
∫
−0.10
1
−kx dx = − k x 2
2
0
−0.10
1
= 0 − − k(−0.01)2 = 0.05 J !
2
Notice that the integral is always this.!
1
Wspring = − k x 2
2
xf
xi
!
It is the integrand that will give you the sign of the work. Here are the possibilities.!
Wspring > 0
x=0
Wspring < 0
Wspring > 0
x=0
Wspring < 0
!
When the mass is moving toward the equilibrium position, the work done by the spring is positive.
The mass speeds up. When the mass is moving away from the equilibrium position, the work
done by the spring is negative. The mass slows down.!
• How fast is the mass traveling after released from rest at the equilibrium position?!
The final speed comes from the work-energy theorem.!
Wtotal = ΔK
1
⇒ Wspring = K f − Ki = K f = mv f2 !
2
1
0.05 J = (0.50)v f2
2
⇒ v f = 0.447 m / s !
page 9
A System of Objects!
Here is what happens when there is more than one object involved.!
Example: Masses and Pulley!
Here is a system of two masses. Let’s say there is no friction.!
m1
m2
!
• After the masses are released from rest and have traveled a distance x, what is the speed of the
masses?!
The total work done on mass 2 is this. Let me call down positive.!
Wtotal,2 = (m2g −T )(x)cos 0° = m2gx −Tx !
The total work done on mass 1 is this. Let me call right positive.!
Wtotal,1 = (T )(x)cos 0° = Tx !
At this point, you can look at the two masses separately.!
1
Wtotal,1 = Tx = ΔK1 = K f ,1 = m1v f2
2
⇒ vf =
1
Wtotal,2 = m2gx −Tx = ΔK 2 = K f ,2 = m2v f2
2
2Tx
!
m1
⇒ vf =
2(m2gx −Tx)
!
m2
Or you can treat the two masses as one bigger system.!
1
Wtotal,system = Wtotal,1 +Wtotal,2 = mgx = ΔKsystem = K f ,system = msystemv f2 !
2
vf =
2m2gx
=
msystem
2m2gx
!
m1 + m2
All of these final velocities are the same. From Newton’s second law and kinematics, the final
velocity of the mass are!
ΣF1 = T = m1a
⇒ a=
ΣF2 = m2g −T = m2a
vf =
T
m1
⇒ Δ(v 2 ) = 2aΔx
⇒ a=
m2g −T
m2
⇒ v f2 = 2
⇒ Δ(v 2 ) = 2aΔx
T
x
m1
⇒ vf =
⇒ v f2 = 2
2Tx
!
m1
m2g −T
x!
m2
2(m2g −T )x
!
m2
Recall that the acceleration is !
a=
m2g
m1 + m2
⇒ Δ(v 2 ) = 2aΔx
⇒ vf =
2m2gx
m1 + m2
page 10