PES 1110 Fall 2013, Spendier
Lecture 20/Page 1
Today:
- General Energy Problems (finish chapter 8)
- Midterm Course Evaluation
Wednesday:
- HW 5 due, HW 6 assigned
- Start chapter 9
No quiz this Friday!
Recap last lecture:
Hooke’s Law:
The force needed to stretch or compress a spring increases linearly with stretching
distance
Fsp = k s
where
s = l – l0
k = spring constant, Unit: N/m
(lower case k)
s = stretching distance
(how far has I been stretched)
l0 = equilibrium length
(spring length when spring is unstretched)
l = length of the spring after force is applied
U el 
1 2
ks elastic potential energy
2
Work the spring does on a mass is saved:
Wel = - ∆ Uel
Today we are going to finish chapter 8. At the end of chapter 8 your book kind of does
this piecemeal of what is happening when I now add gravity and other forces to our
spring problem we discussed last Friday. I want to do them all together at once.
General Energy Problems
The most general problems (this term) involve gravity (mgh), springs (elastic force), and
other forces all doing work. I cannot get better than that – some other force doing work.
For example maybe friction that can do work. Maybe you strapped a rocket to your car
that could do work on your car giving you a very interesting force doing work. We are all
set to deal with it because we know what to do in this case.
To find the total work done, we find the work done by gravity (Wg), by the spring (Wel),
and by the other force (Wother). What this work done by “other force” force mean will be
clear when we do a specific problem.
Wtotal = Wg + Wel + Wother
PES 1110 Fall 2013, Spendier
Lecture 20/Page 2
I always have to go back to the total work, since ALL works together determine the
change on kinetic energy.
Wtotal = ∆ K
We know that
work done by gravity, change in gravitation potential: Wg = - ∆ Ug
work done by a spring, change in elastic potential energy: Wel = - ∆ Uel
After unpacking everything, ∆ stands for final and initial we get:
K  Wtotal  Wg  Wel  Wother  U g  U el  Wother
Using ∆ K = Kf – Ki and so on and after some algebra:
1 2
1
1
1
mvi  mgyi  ksi2  Wother  mv 2f  mgy f  ks 2f
2
2
2
2
Ei + Wother = Efinal
Left hand side: (Initial total energy = initial K energy + initial potential energy due to
gravity + the initial potential energy due to a spring) plus other work done
Right hand side: final total energy = final K energy + final potential energy due to
gravity + the final potential energy due to a spring
It is not a guarantee that we have conservation of energy any more, when we start
throwing in these other forces. Especially, when we have something non-conservative
doing work. But we can still use our basic techniques, our basic ideas behind our
previous discussions, to simply solve for things like how fast and far. Even if the total
energy is changing with time.
This is by for the longest equation of this term but simple to understand. You can use this
equation for all of you energy problems assigned in your next HW. All you need to
remember is that this “Wother” is added on to the initial side of the equation.
PES 1110 Fall 2013, Spendier
Lecture 20/Page 3
Example 1:
An 80 kg man jumps onto a spring platform (k = 18000N/m) going 9m/s. Ignoring
friction, find the maximum compression of the spring, d?
PES 1110 Fall 2013, Spendier
Lecture 20/Page 4
Example 2:
An 80 kg man skydives from a plane 500m above the ground. If he lands with a speed of
4m/s (and was essentially at rest when he jumped), how much work did his parachute do?
The two forces doing work in this example are gravity and the parachute. There are no
springs here:
This “negative” sign makes sense since forces that slow masses down, like the parachute,
do negative work. Wother typically slow things down, so it is a negative value. Therefore
we add it to the left hand side, the initial side, in our equation. When we add Wother, we
are left with less initial energy, which means that we will have less kinetic energy at the
end. Wother will cause the mass to slow down or not to go as far.
Thermal Energy
In most real life problems there will be friction. When we include friction we need to go
back to the statement of conservation of energy, if we include what is called the thermal
energy.
If you rub your hands together, back and forward, your hands feel nice and warm. What
is friction doing? You hands are moving so you have kinetic energy. Friction in the
language of energy is changing the kinetic energy of your rubbing motion in this other
type, heat energy, and called thermal energy.
What is really happening is that the kinetic energy of your hands is transferred into the
molecules inside of your hands.
If you include this thermal energy you can say that work done by friction goes 100 % into
this thermal energy.
Thermal Energy = Eth
PES 1110 Fall 2013, Spendier
Lecture 20/Page 5
So we can write that friction does work – it is the change in thermal energy!
The work done by friction: Wf = −∆Eth
So we can add this back to our equation and we can recover conservation of energy by
including thermal energy:
1 2
1
1
1
mvi  mgyi  ksi2  Wother  W f  mv 2f  mgy f  ks 2f
2
2
2
2
Since Wf = −∆Eth
1 2
1
1
1
mvi  mgyi  ksi2  Wother  Eth  mv 2f  mgy f  ks 2f
2
2
2
2
Rearrange,
1 2
1
1
1
mvi  mgyi  ksi2  Wother  mv 2f  mgy f  ks 2f  Eth
2
2
2
2
The only bad part of this is that I had to be vague about who is acquiring the thermal
energy. I said that all of the kinetic rubbing energy went into my hands. That is not
actually true; you also warm the air around your hands, so thermal energy goes all over
the place. So you cannot really say that the mass gets all of the thermal energy. It is going
out into the surroundings and into the mass.
All I can say is that energy is never destroyed – it goes into some other type of energy!
Example 3:
A 10 kg mass sliding to the right, initially with speed 3m/s, is stopped by friction. How
much thermal energy will be created by this process?