1. 8.4 Changes in Mechanical Energy for Nonconservative Forces

2. Mechanical Energy & Nonconservative Forces
If several forces act (conservative & nonconservative):
The total work done is: Wnet = WC + WNC
WC = work done by conservative forces
WNC = work done by non-conservative forces
The work energy principle still holds:
Wnet = K
For conservative forces (definition of U):WC = -U
 K = -U + WNC
 WNC = K + U
Work done by Nonconservative Forces =
Total change in K + Total change in U

3. Mechanical Energy & Nonconservative Forces, 2
The work done against friction is greater along the red path than along the blue path
Because the work done depends on the path, friction is a nonconservative force

4. Mechanical Energy & Nonconservative Forces, 3
If the work done against the kinetic friction depends on the path taken. As the moves moves from A to B.
W ≡ – ƒkd (8.13)
d is greater for the curve path so W is greater.
The work done by the nonconservative force ƒk is: WNC ≡ –ƒkd
Since: WNC = K + U 

5. Mechanical Energy & Nonconservative Forces, final
In general, if friction is acting in a system:
DEmech = DK + DU = –ƒkd (8.14)
DU is the change in all forms of potential energy
If friction is zero, this equation becomes the same as Conservation of Mechanical Energy

6. Problem Solving Strategies – Nonconservative Forces
Define the isolated system and the initial and final configuration of the system
Identify the configuration for zero potential energy
These are the same as for Conservation of Energy
The difference between the final and initial energies is the change in mechanical energy due to friction

7. Example 8. 6 Motion on a Curve Track No-conservative Forces (Example 8
Example 8.6 Motion on a Curve Track No-conservative Forces (Example 8.7 Text Book)
A child of mass m starts sliding from rest. Frictionless!
Find speed v at the bottom.
DEmech = DK + DU
DEmech =(Kf – Ki) + (Uf – Ui) = 0
(½mvf2 – 0) + (0 – mgh)= 0
½mvf2 – mgh = 0 
Same result as the child is falling vertically trough a distance h!

8. Example 8.6 Motion on a Curve Track, final
If a kinetic friction acts on the child, find DEmech
Assuming: m = 20.0 kg and vf = 3.00m/s
DEmech = (Kf – Ki) + (Uf – Ui) ≠ 0
DEmech = ½mvf2 – mgh
DEmech = ½(20)(3)2 – 20(9.8)(2) = –302J
If we want to find k:
DEmech = –302J
DEmech = –ƒkd = –knd = –302J 
k= 302/nd

9. Example 8. 7 Spring-Mass Collision No-conservative Forces (Example 8
Example 8.7 Spring-Mass Collision No-conservative Forces (Example 8.9 Text Book)
Frictionless!
K +Us = Emech remains constant
Assuming: m= 0.80kg vA = 1.2m/s k = 50N/m
Find maximum compression of the spring after collision (xmax)
EC = EA  KC + UsC = KA + UsA
½mvC2 + ½kxmax2 = ½mvA2 + ½kxA2
0 + ½kxmax2 = ½mvA 

10. Example 8.7 Spring-Mass Collision, 2
If friction is present, the energy decreases by: DEmech = –ƒkd
Assuming: k= m= 0.80kg vA = 1.2m/s k = 50N/m
Find maximum compression of the spring after collision xC
DEmech = –ƒk xC = –knxC
DEmech = –kmgxC 
DEmech = –3.92xC (1)

11. Example 8.7 Spring-Mass Collision, final
Using: DEmech = Ef – Ei
DEmech = (Kf – Uf) + (Ki – Ui)
DEmech = 0 – ½kxC2 + ½mvA2 + 0
DEmech = – 25xC (2)
Taking: (1) = (2)
– 25xC = –3.92xC
Solving the quadratic equation for xC :
xC = 0.092m < 0.15m (frictionless)
Expected! Since friction retards the motion of the system
xC = – 0.25m
Does not apply since the mass must be to the right of the origin.

12. Example 8.8 Connected Blocks in Motion N-nconservative Forces (Example 8.10 Text Book)
The system consists of the two blocks, the spring, and Earth. Gravitational and potential energies are involved
System is released from rest when spring is unstretched.
Mass m2 falls a distance h before coming to rest.
Find k

13. Example 8.8 Connected Blocks in Motion, 2
The kinetic energy is zero if our initial and final configurations are at rest
Block 2 undergoes a change in gravitational potential energy
The spring undergoes a change in elastic potential energy
Emech = K + Ug + US
Emech = Ug + US
Emech = Ugf – Ugf + Usf – Usi
Emech = 0 – m2gh + ½kh2 – 0
Emech = – m2gh + ½kh2 (1)

14. Example 8.8 Connected Blocks in Motion, final
If friction is present, the energy decreases by:
DEmech = –ƒkh = – km1gh (2)
Taking (1) = (2)
– m2gh + ½kh2 = – km1gh
Solving for k
km1gh = m2gh – ½kh2 
k = (m2gh)/ (m1gh) – (½kh2)/(m1gh) 
k = m2/m1 – (½kh)/m1g
This is another way to measure k !!!

15. We will not cover Section: 8.5 Please Read it!!! Here you have a taste
8.5 Conservative Forces and Potential Energy
NOTE:
We will not cover Section: 8.5
Please Read it!!!
Here you have a taste

16. Conservative Forces and Potential Energy
Define a potential energy function, U, such that the work done by a conservative force equals the decrease in the potential energy of the system
The work done by such a force, F, is
(8.15)
DU is negative when F and x are in the same direction

17. Conservative Forces and Potential Energy
The conservative force is related to the potential energy function through
(8.15)
The x component of a conservative force acting on an object within a system equals the negative of the potential energy of the system with respect to x

18. Conservative Forces and Potential Energy – Check
Look at the case of a deformed spring
This is Hooke’s Law

19. 8.6 Energy Diagrams and Equilibrium
Motion in a system can be observed in terms of a graph of its position and energy
In a spring-mass system example, the block oscillates between the turning points, x = ±xmax
The block will always accelerate back toward x = 0

20. Energy Diagrams and Stable Equilibrium
The x = 0 position is one of stable equilibrium
Configurations of stable equilibrium correspond to those for which U(x) is a minimum
x =xmax and x = –xmax are called the turning points

21. Energy Diagrams and Unstable Equilibrium
Fx = 0 at x = 0, so the particle is in equilibrium
For any other value of x, the particle moves away from the equilibrium position
This is an example of unstable equilibrium
Configurations of unstable equilibrium correspond to those for which U(x) is a maximum

22. Neutral Equilibrium
Neutral equilibrium occurs in a configuration when over some region U is constant
A small displacement from a position in this region will produce either restoring or disrupting forces

23. Potential Energy in Molecules
There is potential energy associated with the force between two neutral atoms in a molecule which can be modeled by the Lennard-Jones function

24. Potential Energy Curve of a Molecule
Find the minimum of the function (take the derivative and set it equal to 0) to find the separation for stable equilibrium
The graph of the Lennard-Jones function shows the most likely separation between the atoms in the molecule (at minimum energy)

25. Force Acting in a Molecule
The force is repulsive (positive) at small separations
The force is zero at the point of stable equilibrium
The force is attractive (negative) when the separation increases
At great distances, the force approaches zero

26. Material for the Final Examples to Read!!! Example 8.6 (page 230)
Homework to be solved in Class!!!
Questions: 13, 21
Problems: 36, 45, 48