Transcript Potential Energy and Conservation of Mechanical Energy

The power of Niagra Falls
Height: 167 ft
Flow: 600,000 U.S. gallons per second
The power of Einstein
Kinetic energy: E = ½ mV2
Energy of matter: E = mc2
Potential Energy and Conservation of
Mechanical Energy
Chapter 8-1 / 8-3
CONSERVATIVE FORCES are forces that…..
•Produce only mechanical motion
•Store energy in mechanical motion
Examples are gravity and spring forces.
NON-CONSERVATIVE FORCES are forces that….
•Create energy in the form of heat, sound, or other
non-mechanical process
•Cause a transfer of energy from one system to
another
Example is friction.
Work of a conservative force can be positive or negative.
W=F*D
Dx
Dx
F
Work of gravity is NEGATIVE.
F
Work of gravity is POSITIVE.
Work done by a conservative force around a
closed path is zero.
This leads to an important conclusion…..
The work done by a conservative force is
independent of the path, and depends only on the
starting and ending points.
Pick any starting and ending points.
Closed path, W=0.
A
W2
W1
B
A
W1 = WAB
W2 = WBA
W 1 + W2 = 0
W 1 + W3 = 0
W3
W1
B
So, all paths from B to A take the
same amount of work.
Conservative forces produce
“Potential Energy”
W = - DU = -(Ufinal - Uinitial)
W = FDx = -MgH
Ufinal
Dx = H
F
so
MgH = (Ufinal - Uinitial)
Mgyfi - Mgyin = (Ufinal - Uinitial)
Uinitial
F
Gravitational Potential Energy
U = Mgy
Which graph of potential energy describes the
action of the force in the picture below?
To solve, this, note:
1. Picture 1
2. Picture 2
3. Picture 3
W   DU
W  F Dx
DU
F 
Dx
Work, Potential Energy, and
Mechanical Energy
W = - DU = -(Ufinal - Uinitial)
and
W = DK
So,
Work-energy theorem
W – W = 0 = DK + DU
Leads to the Conservation of Mechanical Energy
E=K+U
E is CONSTANT, SO
DE = 0
Conservation of energy can simplify problem solving.
Block dropped from height H. What is
speed of block just before impact with
ground?
Ui = MgH
Ki = 0
Ki = ½ M v2
Uf = 0
E = constant
E=K+U
H
Initial
Final
1
Mv 2  0
2
v 2  2 gH
0  MgH 
v  2 gH
Path doesn’t matter!
Initial
Final
1
0  MgH  Mv 2  0
2
v 2  2 gH
v  2 gH
Potential energy of a spring
If a spring is
COMPRESSED or
STRETCHED and
amount DX from its
equilibrium position, it has
a stored energy, equal
to…
1
2


U  k Dx
2
This is the same value that you have for the Work done on the spring.
Pinball shooter
An in-class problem
solving exercise.
DX
M
V
A spring is initially compressed by an amount DX by a mass M. The
mass is released and slides without friction. Given the spring constant
K, the compression distance, and the mass M, what is a formula for
the final velocity of the block?
How to solve:
Write down the initial potential and kinetic energy. Next write
down the final potential and kinetic energy. Set them equal
(conservation of mechanical energy). Solve for V.
Pinball shooter: step by step
DX
Q: Find a formula for final velocity.
M
V
1. Write down initial potential and kinetic energy
Initial Potential Energy: U = ½ K DX2
Initial Kinetic Energy: K = ½ M V2 = 0
2. Write down the final potential energy and kinetic energy.
3. Set the initial and final energy to be equal (conservation of energy).
4. Solve for V.
Pinball Shooter: The final velocity of the
block is given by ……
DX
M
V
GIVEN: DX, M, K, g
1. V = sqrt (2g DX)
2. V = DX sqrt (K/M)
3. V = ½ K (DX)^2
Pinball Shooter: The final velocity of the
block is given by ……
Second Chance! Work with your Neighbor.
DX
M
V
GIVEN: DX, M, K, g
1. V = sqrt (2g DX)
2. V = DX sqrt (K/M)
3. V = ½ K (DX)^2
Work done by a spring.
From point A to B, spring is pulling,
work is negative.
F
DX
C
From point C to point A, spring is
pulling, work is positive.
What is the work done BY THE SPRING?
Path #1:
F
DX
1
1
1
1
2
2
W   k (4) 2   k (4) 2  k 2    k 2 
2
2
2
2

Path #2:
1
1
1
1
2
2
2
W   k (2) 2  k 2   k 2   k  2 
2
2
2
2
Tricky part: moving
from 4 cm to 2 cm.
Compare to W=-DU
method.
Springs and gravity: potential
stored in spring.
At equilibrium,
Force of spring is equal to force of gravity.
KL = Mg
Potential energy stored in spring:
U = ½ K L2
L
M
Potential energy of mass and spring together
A different problem from the previous slide.
Since only DIFFERENCES in potential energy
matter, we can define the ZERO of potential to
be the equilibrium position of the mass, after it
stretches the spring.
Potential of mass: mgy
Potential of spring: ½ Ky2
Total: U = mgy + ½ Ky2
Pulleys: solve using energy conservation
E init = 0 (my choice of potential)
1
1
m1v 2  m2 v 2  m1 gh  m2 gh
2
2
0
E FINAL 
1
m1  m2 v 2  m2  m1 gh
2
CHECK: Suppose m1 was equal to zero? What if masses are equal?
Combo problem: spring and falling body.
M
H
L
Mass M drops from a height L onto a spring
loaded platform. How much does the spring
compress? Spring constant is K.
Use conservation of energy.
Write down initial energy of mass and
spring.
(Be sure to use an easy definition of initial
energy.)
Write down final energy of mass and spring.
This will be at point of maximum
compression, when the mass STOPS
MOVING!
M
H
L
How much does the spring
compress?
Solve this equation:
1. ½ KL^2 = MgL
2. ½ KL^2 = Mg(L+H)
3. ½ KL^2 = sqrt(MgH)