Download 6-3 Work and Energy in 3-D

6-3 Work and Energy in 3-D
1st – complete the 6.3 Warm-up Exercise 
We learned previously that Fnet ∙ v = dK/dt, where K = ½ mv2. To convert this into 3-D,
integrate both sides of the equation with respect to time, and you get:
Since ds = v dt, where ds is the displacement during time dt, and because (dK/dt)dt = dK,
you can rewrite this as
Q: What does the integral on the left give you?
A: Total work done on the particle, Wtotal
This is the proof of the work-energy theorem in 3-D using calculus!
See Ex 6-12, p 166-167
You and a friend are at a ski resort with two ski runs (kind of lame-oh ), a beginner’s
run and an expert run. Both runs begin at the top of the lift and end at the same finish
line at the bottom. Let h be the vertical descent for both runs. The beginner’s run is
(obviously) less steep, thus longer than the expert run. You and your friend, who is a
much better skier than you, are testing some experimental frictionless (really? ) skis. To
make things interesting, you wager that her speed at the finish line will not be greater
than yours, and she accepts. The conditions of the bet are that you both start from rest at
the top of the lift, and you both coast the entire time. Assuming no air drag (really? ),
who wins the bet?
NOTE: Since you are both coasting, you can assume you are moving like particles. (If
not, you couldn’t use the work-energy theorem!)
Q: Which forces are acting on you? Of these forces, which contribute to your motion?
A: Gravity and Fn; Both
Wtotal = ½ mvf 2 - ½ mvi 2
Wtotal = Wn + Ws
dWn = Fn ∙ ds = Fn cos Ø ds
The force on you, mg, is constant, but Fn is not! Why?
Ø = 90º
dWn = Fn cos(90º) ds = 0, so…
This means that the work done by the normal force is zero. Should it be?
Yes! Though the normal is constantly changing direction, it is always  to the direction of
motion (ds)  no work is done in that direction - EVER.
Wg = mg · s = -mg j · (Δxi + Δyj)
Wg = -mg Δy
Δy = -h
Wg = mgh
Wn + Wg = ΔK
0 + mgh = ½ mvf 2 – 0
vf = √2gh
YOU WIN!!!
Woot!
The bet was that she would not be going
faster!
She will cross the finish line first, but that wasn’t the bet! This problem
proved that work due to gravity, mgh, is independent of the steepness of the
hill, or the length of the path.
A bit morr on this to come in a moment. Study this example and it’s endresult!
6-4 Potential Energy
Q: How do “total work” and KE relate?
A: Total work done on a particle = the change in KE
Sometimes, the work done by a system does not increase the total KE of a system, but
rather, is stored as PE for the entire system.
If you lift a barbell to a height, h, then the work done by gravity on the barbell is – mgh.
If the barbell starts at rest and ends at rest, ∆ KE = 0. If ∆ KE = 0, then no work was
done on the system, according to definition. What that means is that the work done by
your hands is +mgh. If the earth and barbell are a system of two particles (the lifter is not
part of the system), then the external forces on the earth-barbell system are:
1. the gravitational attraction the lifter exerts on the earth
2. the force his feet exert on the earth
3. the force, mg, he exerts on the barbell
So…, the total work done on the earth-barbell system by all the forces external to the
system is mgh. This work gets stored as PE. Really, this is called gravitational PE.
Springs store PE in much the same way as the lifter-barbell problem in that work is not
expressed in terms of KE, but rather as stored PE.
Conservative Forces
If you ride a chairlift up a hill, the work done by gravity on you is – mgh, and the work
done by the lift on you is +mgh. The total work done by gravity on you for the round trip
= 0, independent of the path you take. This is what is meant by a conservative force.
If you consider the “you-earth” system to be a two particle system (ignore the lift), the
work done on you to raise you to the top of the hill is stored as gravitational PE. As you
ski down, it is converted to KE of your motion.
So, for Ex 6-12 above, not only does it typify a conservative force (gravity), but also proves
that the end velocity for the conservative gravitational force is independent of path.
Examples of Conservative Forces:
 Gravitational force
 Spring force
 Electromagnetic force
These are the forces for which there is a defined potential energy.
PE Functions
Q: HOW does the potential energy of an object change?
A: By incurring a change of position.
Q: WHY does potential energy change?
A: Because some force has acted upon it!
Work done by a conservative force does not depend on the path, only on the endpoints.
So, we can define a PE function, U as one associated with conservative force. Because as
you ski downhill, for example, the work done by gravity decreases the PE of the system,
we can define the PE function accordingly:
For infinitesimal displacements, we have
If dU = - F∙ds = - (-mgj) ∙ (dxi + dyj + dzk) = + mg dy
Integrating gives you
Where U0 = the arbitrary constant of integration.
Q: Is the actual value of U important?
A: Nope. Only a change in PE is defined.
See Example 6-13, p 169-170
A bottle of mass 0.350 kg falls from rest from a shelf
1.75 m above the floor. Find the PE of the bottleearth system when the bottle is on the shelf and just
before impact with the floor. Find the KE just
before impact with the floor.
PE of a Spring
Springs are a type of conservative force.
dU = - F∙ds = - Fxdx = - (-kx)dx = + Kxdx, so for a spring, and choosing U0 = 0,
See Example 6-14, p 171
A system consists of a basketball player, the rim of a basketball hoop, and the earth.
Assume PE = 0 when the player is standing on the floor and the rim is horizontal. Find
the total PE when the player hangs from the rim. Also, assume the player can be
described as a point mass of 110 kg at 0.8 m above the floor when hanging. The force
constant of the rim is 7.2 kN/m and the front of the rim is displaced 15 cm.
Non-conservative Forces
Q: Are all closed-curve motion paths conservative?
A: Nope. If you can prove that the total work done is not = 0, then the closed-curve path
is not conservative.
Basically speaking, the minute you prove that the total work done is not = 0, then the
closed-curve path is not conservative. But you have to be careful; if you do state that the
closed-curve path is conservative, then it has to be conservative for all of the paths that get
you from the starting to the ending point.
NOTE: If a path is non-conservative, no PE function can be defined for it.
Some Examples of Non-Conservative Forces:
 Kinetic frictional force
 Drag force
For example, if you send a block sliding across a non-frictionless floor, the kinetic friction
from the floor does negative work on the block. The block, then, slows down by
transferring its kinetic energy to thermal energy of the atoms / molecules involved. Since
we cannot transfer this thermal energy back into kinetic frictional energy, this acquired
thermal energy is not a potential energy for this system. Thus, these forces are not
conservative.
PE and Equilibrium
For any general conservative force in one dimension (x, for example), dU = - Fds = - Fxdx
So, the force is the negative derivative of the PE function.
If we’re talkin’ springs, then U = ½ kx2 and we get Fx = - dU / dx = - (1/2 kx2) = - kx
Q: For the diagram below which shows a plot of U = ½ kx2 versus x for a block and
spring, how would you represent the derivative of the function?
A: As you KNOW, it’s the slope of the tangent to the curve! 
Q: What is the force equal to?
A: The negative of the slope of the curve
Q: What happens at x = 0?
A: Fx = - dU/dx is zero, and the block is in equilibrium. That’s not new news!
Q: When x is positive, the slope is ___________and the force is _____________.
When x is negative, the slope is ___________and the force is _____________.
A: + - - + 
Q: What do we know about the accelerations at any given point?
A: In either case, the force is in the direction that will accelerate the block back to a lower
PE.
What that really means is that if x varies a bit from x = 0, the force will be directed back to
x = 0. The equilibrium that happens at x = 0 is stable because any small displacement
results in a restoring force that will accelerate the particle back to its equilibrium position.
Figure 6-28
Take a look at figure 6-28, p 173.
Q: What is happening here? Will this result in a stable equilibrium? Why / why not?
A: Unstable – concave down will send the little particle merrily on its way away from
the relative maximum. You find it by taking d2U/dx2
Take a look at Figure 6-29, p 173. What
is happening here? Will this result in a
stable equilibrium? Why / why not?
Figure 6-29
So, one morr time – when the slope of the PE graph is zero, the object is in some sort of
equilibrium.
NOTE: Taking the second derivative will give you the point of inflection, thus telling you
whether your point is a max or min, and therefore telling you the state of your
equilibrium!
Types of Equilbria and Their Relationship to the second derivative of the graph of U:
Slope d2U/dx2 Equilibrium
exists?
Type of
Object Stays
Equilibrium in Equilibrium?
0
+
Y
Stable
Y
0
-
Y
Unstable
N
0
0
Y
Neutral
Y
≠0
N
See Example 6-15, p 173 – 174
In the region –a < x < a, the force on a particle is represented by the PE function
1
1 

U = - b 
 where a and b are positive constants. Find
a x
a x
a) the force Fx
b) at what value of x the force = 0
c) at the location where force = 0, is the equilibrium stable or unstable?
Study that summary on p 174 – 175.
EX Morr 

The diagram below shows three paths connecting points a and b. A single force F does
the indicated work on the particle moving along each path in the indicated direction. On
the basis of this information, is the force conservative?
- 60 J
a
60 J
b
60 J
A: Nope. Look at the round trip for the smaller loop. Total work ≠ 0.