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The power of Niagra Falls
Height: 167 ftFlow: 600,000 U.S. gallons per second
The power of EinsteinKinetic 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.
x
x
F F
Work of gravity is NEGATIVE. Work of gravity is POSITIVE.
W = F * D
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.Closed path, W=0. Pick any starting and ending points.
A
B
A
B
W1
W2
W1 = WAB
W2 = WBA
W1 + W2 = 0W3
W1 + W3 = 0
So, all paths from B to A take the same amount of work.
W1
Conservative forces produce “Potential Energy”
x = H
F
F
Uinitial
Ufinal
W = - U = -(Ufinal - Uinitial)
W = Fx = -MgH
so
MgH = (Ufinal - Uinitial)
Gravitational Potential EnergyU = Mgy
Mgyfi - Mgyin = (Ufinal - Uinitial)
Which graph of potential energy describes the action of the force in the picture below?
1. Picture 12. Picture 23. Picture 3
To solve, this, note:
x
UF
xFW
UW
Work, Potential Energy, and Mechanical Energy
W = - U = -(Ufinal - Uinitial)
and
W = K Work-energy theorem
Leads to the Conservation of Mechanical Energy
E = K + U
E is CONSTANT, SOE = 0
So,W – W = 0 = K + U
Conservation of energy can simplify problem solving.
H
Block dropped from height H. What is speed of block just before impact with ground?
Ui = MgH Ki = 0
E = constantE = K + U
Uf = 0 Ki = ½ M v2
gHv
gHv
MvMgH
2
2
02
10
2
2
Initial Final
Path doesn’t matter!
gHv
gHv
MvMgH
2
2
02
10
2
2
Initial Final
Potential energy of a spring
If a spring is COMPRESSED or STRETCHED and amount X from its equilibrium position, it has a stored energy, equal to…
22
1xkU
This is the same value that you have for the Work done on the spring.
Pinball shooter
XM
V
A spring is initially compressed by an amount X 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.
An in-class problem solving exercise.
Pinball shooter: step by step
X MV
Q: Find a formula for final velocity.
Initial Potential Energy: U = ½ K X2
1. Write down initial potential and kinetic energy
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 ……
1. V = sqrt (2g X)2. V = X sqrt (K/M)3. V = ½ K (X)^2
X MV
GIVEN: X, M, K, g
Pinball Shooter: The final velocity of the block is given by ……
1. V = sqrt (2g X)2. V = X sqrt (K/M)3. V = ½ K (X)^2
X MV
GIVEN: X, M, K, g
Second Chance! Work with your Neighbor.
Work done by a spring.
What is the work done BY THE SPRING?
From point A to B, spring is pulling, work is negative.
FX
C
From point C to point A, spring is pulling, work is positive.
F
XPath #1:
Path #2:
2222 22
12
2
12
2
1)2(
2
1 kkkkW
2222 22
12
2
1)4(
2
1)4(
2
1kkkkW
Tricky part: moving from 4 cm to 2 cm.
Compare to W=-U method.
Springs and gravity: potential stored in spring.
M
L
At equilibrium,Force of spring is equal to force of gravity.
KL = Mg
Potential energy stored in spring:
U = ½ K L2
Potential energy of mass and spring togetherA 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)
02
1
2
121
22
21
ghmghmvmvmEFINAL
ghmmvmm 122
212
1
CHECK: Suppose m1 was equal to zero? What if masses are equal?
Combo problem: spring and falling body.
H
L
M 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!
How much does the spring compress?
1. ½ KL^2 = MgL2. ½ KL^2 = Mg(L+H)3. ½ KL^2 = sqrt(MgH)
H
L
M
Solve this equation: