1. Define Potential Energy of a Hooke’s Law spring Use Energy diagrams
Lecture 15
Goals:
Chapter 10
Define Potential Energy of a Hooke’s Law spring
Use Energy diagrams
Chapter 11 (Work)
Define Work
Employ the dot product
Heads up:
Exam 2 7:15 PM Thursday, Nov. 3th 1

2. Energy for a Hooke’s Law springm
Associate ½ ku2 with the “potential energy” of the spring

3. Energy for a Hooke’s Law springm
Ideal Hooke’s Law springs are conservative so the mechanical energy is constant

4. Energy diagrams Ball falling Spring/Mass system Emech Emech y
In general:
Ball falling
Spring/Mass system
Emech
Energy K y U
Emech K
Energy U
u = x - xeq

5. Energy (with spring & gravity)1 3 2 h -x
mass: m
Given m, g, h & k,
how much does the spring compress?
Emech = constant (only conservative forces)
At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0
Em1 = Ug1 + Us1 + K1 = mgh
Em2 = Ug2 + Us2 + K2 = ½ mv2
Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx

6. Energy (with spring & gravity)1 3 2 h -x
mass: m
Given m, g, h & k, how much does the spring compress?
Emech = constant (only conservative forces)
At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0
Em1 = Ug1 + Us1 + K1 = mgh
Em2 = Ug2 + Us2 + K2 = ½ mv2
Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx
Given m, g, h & k, how much does the spring compress?
Em1 = Em3 = mgh = -mgx + ½ kx2 

7. Energy (with spring & gravity)1 3 2 h -x
mass: m
Given m, g, h & k, how much does the spring compress?
Emech = constant (only conservative forces)
At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0
Em1 = Ug1 + Us1 + K1 = mgh
Em2 = Ug2 + Us2 + K2 = ½ mv2
Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx
Given m, g, h & k, how much does the spring compress?
Em1 = Em3 = mgh = -mgx + ½ kx2  Solve ½ kx2 - mgx - mgh = 0

8. Energy (with spring & gravity)1
mass: m 2 h 3 -x
When is the child’s speed greatest?
(A) At y = h (top of jump)
(B) Between h & 0
(C) At y = 0 (child first contacts spring)
(D) Between 0 & -x
(E) At y = -x (maximum spring compression)

9. Energy (with spring & gravity)1 2 h mg 3 kx -x
When is the child’s speed greatest? (D) Between y2 & y3
A: Calc. soln. Find v vs. spring displacement then maximize
(i.e., take derivative and then set to zero)
B: Physics: As long as Fgravity > Fspring then speed is increasing
Find where Fgravity- Fspring= 0  -mg = kxVmax or xVmax = -mg / k
So mgh = Ug23 + Us23 + K23 = mg (-mg/k) + ½ k(-mg/k)2 + ½ mv2
 2gh = 2(-mg2/k) + mg2/k + v2  2gh + mg2/k = vmax2

10. Chapter 11: Energy & Work
Impulse (Force over a time) describes momentum transfer
Work (Force over a distance) describes energy transfer
Any single acting force which changes the potential or kinetic energy of a system is said to have done work W on that system
DEsys = W
W can be positive or negative depending on the direction of energy transfer
Net work reflects changes in the kinetic energy
Wnet = DK
This is called the “Net” Work-Kinetic Energy Theorem

11. Work performs energy transfer
If only conservative forces present then work either increases or decreases the mechanical energy.
It is important to define what constitutes the system

12. How much work is done after the ball makes one full revolution?
Circular Motion
I swing a sling shot over my head. The tension in the rope keeps the shot moving at constant speed in a circle.
How much work is done after the ball makes one full revolution? v
(A) W > 0 Fc
(B) W = 0
(C) W < 0
(D) need more info

13. Examples of “Net” Work (Wnet)
DK = Wnet
Pushing a box on a smooth floor with a constant force; there is an increase in the kinetic energy
Examples of No “Net” Work
DK = 0 = Wnet
Pushing a box on a rough floor at constant speed
Driving at constant speed in a horizontal circle
Holding a book at constant height
This last statement reflects what we call the “system”

14. Again a constant F along a line (now x-dir)
Recall eliminating t gives
Multiply by m/2
And m ax = Fx (which is constant)

15. Constant force along x, y & z directions
Then for each x y z component Fx Fy Fz there is work
W = Fx Dx + Fy Dy + Fz Dz
Notice that if  is the angle between F and :
|F| cos  is the compoent of F parallel to Dr
|F| cos  |Dr| = W = Fx Dx + Fy Dz + Fz Dz
So here we introduce the “dot” product
F  Dr ≡ |F| cos  |Dr| = Fx Dx + Fy Dz + Fz Dz = DK = Wnet
A Parallel Force acting Over a Distance does Work F q
F|| Δr

16. Scalar Product (or Dot Product)î A Ax Ay q
Useful for finding parallel components
A  î = Ax
î  î = 1
î  j = 0
Calculation can be made in terms of components.
A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )
Calculation also in terms of magnitudes and relative angles.
A  B ≡ | A | | B | cos q
You choose the way that works best for you!

17. Scalar Product (or Dot Product)
Compare:
A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )
with A as force F, B as displacement Dr
Notice if force is constant:
F  Dr = (Fx )(Dx) + (Fy )(Dz ) + (Fz )(Dz)
Fx Dx +Fy Dy + Fz Dz = DK
So here
F  Dr = DK = Wnet
Parallel Force acting Over a Distance does Work

18. Net Work: 1-D Example (constant force)
A force F = 10 N pushes a box across a frictionless floor for a distance x = 5 m.
Start
Finish F
q = 0° x
Net Work is F x = 10 x 5 N m = 50 J
1 Nm ≡ 1 Joule and this is a unit of energy
Work reflects energy transfer

19. For Wednesday: Read Chapter 12, Sections 1-3, 5
Recap
Next time:
Power (Energy / time)
Start Chapter 12
Assignment:
HW7 due Tuesday, Nov. 1st
For Wednesday: Read Chapter 12, Sections 1-3, 5
do not concern yourself with the integration process 1