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Chapter 8: Potential Energy
8.1: Potential Energy of a System
There are many types of stored potential energy; below are some examples.
1. Compressed energy (an inflated balloon that escapes your hand before tying, compressed springs, etc)
2. Chemical energy (batteries)
3. Electrostatic (in the form of charges, or capacitors)
4. Magnetic fields (inductors)
5. Nuclear binding energy (weak and strong forces in the nucleus).
This chapter deals primarily with gravitational potential energy of Energy of
(vertical)
Position.
We know if we exert energy to raise a 100 N dog to a height of 1 meter above the floor, we have done
100 Nm of work on the dog (at the same time gravity has done -100 Nm on the object which means NO net wok is done.)
Then if we drop this dog from the height of 1 meter, the dog will accelerate
and will delivery 100 Nm of energy to the floor.
(down)
through one meter
Immediately before the dog lands on the floor the dog will be moving at maximum speed.
In other words the energy of POSITION has now transferred to energy of MOTION.
Why did I use a dog for an example? Do you think that a 100 Newton rock will damage the ground more than a 100
Newton dog? This is actually an introduction for Ch 9…we’ll revisit this example in Chapter 9.
8.2: The Isolated System
Lecture book Examples:
8.3 The Pendulum
8.2 Ball in Free Fall - Conservation of mechanical
energy, isolated system, elastic potential energy
8.4 A Grand Entrance
8.5 The Spring-Loaded Popgun
How fast is the dog moving immediately before
impact?
I applied a 100 N force through 1 meter, so I did 100
Nm of work to raise the dog. Gravity did -100 Nm of
work.
What is a non-conservative force? Friction is generally
the most common non-conservative force. At the
beginning of the class (several weeks ago) we discussed
when air friction on a car really comes into play…we
decided that was between 40 and 60 mph (or 20 to 30
m/s).
The 100N dog is now 1 meter above the floor. The
energy of position is 100 Nm: weight of dog (mg) time
distance the dog will fall before reaching the floor.
Do you think the dog will be traveling 60 mph before
hitting the ground if I drop him from one meter?
Obviously not…so we can easily ignore air friction in this
problem.
If no non-conservative forces are acting during this
time, we know that ALL of the energy of Position has
transferred to energy of Motion.
As discussed in Chapter 5…we always ignore air friction
in the class unless otherwise instructed.
Review
F
F
F
F
∫Fdx
F(xf – xi)
FΔx
Work
=m a
= m dv/dt
Apply chain rule
= m dv/dx (dx/dt)
= m dv/dx
v
= m ∫v dv if F not dependant on position
= m ½(vf2 – vi2)
= Δ½mv2
= ΔKE
So if all of the energy of position has converted
Why do we know it’s Kinetic Energy?
Ans: we know the units of work and energy is Nm
{Newton meters (or a Joule)} and so it must be energy
of motion since it’s velocity squared term.
T  TE or Total Energy
K  KE or Kinetic Energy
U  Potential Energy
Energy losses are simply energy transferred out of the
to energy of motion with no losses then
mghmax
= ½mvmax2
v2
= 2gh
2
vdog = 2(10m/s2)(1 m)
vdog = 4.5 m/s
Work = ∫-kx dx
Work = -k∫x dx
Work = -k ½(xf2 – xi2)
Wspring = ½kxi2 – ½kxf2
system and are unusable for the system
What if F is dependant on position?
Work = ∫F dx
for example Hooke’s Law: F = -kx
Example
Work = ½kxi2 - ½kxf2
A spring requires 0.09 Nm of work to 0.09 = ½ k .032
pull it back 3 cm. You pull the spring k = 200 N/m
back 10 cm. What is the exit velocity
of a 50 gram block from this
horizontal spring?
Demo: Hopper Popper: ME-HAR
As I deform the popper (inside out) and release it,
what happens?
...it POPs UP. The stored elastic PE is released.
Is this height important?
While everyone was taking the quiz, I popped the
Hopper Popper from the floor...and it rose about
Wsping
½ k(0)2 - ½ k(.1)2
0
- ½k(.1)2
0
-½ 200(.1)2
v = (40)½
v = 6.3 m/s
= ΔKE
= ΔKE
= -½ mv2
= -½ .05 v2
(A) So how much Elastic Potential energy is stored in
the Hopper Popper
(A-ans)
Should we use Us = ½kx²?
This isn't exactly a spring so we don't know if this is
valid...especially since "x" would be very difficult to
measure.
So how do we measure it?
7.5 feet.
Remember...I tried it from the table and it hit the
ceiling.
Demo: Track, Conservation of
Energy: ME-M-TC
As the Popper is rising Gravity is doing work
work) upon the Popper, thus
Work by gravity = mg(hmax- hmin)
(or negative
I'm estimating the mass of the Popper to be about 35
grams.
Work = 0.035kg*9.8m/s²*2.4m
Work = 0.84 Nm
Work = 0.84 Joules
8.3: Conservative and Non-Conservative Forces
Below, the spring (k = 10,000
N/m) is stretched by 10 cm and
the 2kg mass is propelled up the
ramp. “L” is the point at which
the 2kg object stops. If μ =
0.15, what is the length of L?
The spring propels the block up the ramp. As the block leaves the
spring KEmax is achieved. The KE is transferred to PEgrav and is
reduced by friction as it rises up the ramp?
NOTE: Ff through a distance reduces KE
Wspring = ½kxi2 – ½kxf2
Wspring = ΔK
ΔK
–
Ff d
½kx2
- μ cos 30°mg L
½10000(.1)2 – 0.15 0.866 (2)(1)10
50 –
2.6 L
L = 4.0 m
Ff = μ N
Ff = μ cos 30°mg
= mgh
= mg L sin 30°
L = 2(10) L ½
= 10 L
Demo: Pendulum, Bowling Ball: ME-M-PB
sin 30° = h / L
h = L sin 30°
8.4: Changes in Mechanical Energy for Non-Conservative Forces
Key Concepts: Change in mechanical energy of a
system, due to friction within the system
Examples:
8.6 Crate Sliding Down a Ramp
8.7 Motion on a Curved Track
8.8 Let's Go Sking!
8.9 Block-Spring Collision
8.10 Connected Blocks in Motion
8.5: Relationship Between Conservative Forces and Potential Energy
Tarzan & Jane
Jane whose mass is 40 kg, needs to swing across a
river of width 60 meters filled with man-eating
crocodiles to save Tarzan (70 kg) from danger.
However, she must swing into a wind exerting a
constant horizontal force F on a vine having a length
50 meters, and initially making an angle 50° with the
vertical. (a) with what minimum speed must Jane
begin her swing to just make it to the other side?
(b) Once the rescue is complete, Tarzan and Jane
must swing back across the river. With what
minimum speed must they begin their swing?
Workgravity - Workwind = ΔKE
Workgravity = m g h
Workgravity - Workwind
= ½m vf² - ½m vi²
5000 Nm – 6000 Nm = 0 - ½40kg vi²
vi = 7.1 m/s
= mg (50 cos26° – 50 cos50°)
= mg(44.9 - 32.1)
= 5000 Nm
Workwind = Fwind ∫dr
= 100 N * 60 m
= 6000 Nm
(b) This time Air is aiding Jane and Tarzan's crossing
ΔKE + Workwind =
Workgravity
½mv² + 6000
= (40 + 70)(g)(12.8 m)
½110kg v²
= 7800 Nm
v = 11.9 m/s
8.6: Energy Diagrams and Equilibrium of a System
stable equilibrium, unstable equilibrium, neutral equilibrium
Examples:
8.11 Force and Energy on an Atomic Scale