1. Until recently… 1

2. Lecture 14 Goals: Chapter 10
Understand the relationship between motion and energy
Define Potential Energy in a Hooke’s Law spring
Exploit conservation of energy principle in problem solving
Understand spring potential energies
Use energy diagrams
Chapter 11
Understand the relationship between force, displacement and work
Assignment:
HW6 due Tuesday Oct. 26th
For Monday: Read Ch. 11 1

3. -mg (yf – yi) = ½ m ( vyf2 -vyi2 )
Energy
-mg Dy= ½ m (vy2 - vy02 )
-mg (yf – yi) = ½ m ( vyf2 -vyi2 )
A relationship between
y- displacement and change in the y-speed squared
Rearranging to give initial on the left and final on the right
½ m vyi2 + mgyi = ½ m vyf2 + mgyf
We now define mgy = U as the “gravitational potential energy”

4. Energy ½ m vi2 + mgyi = ½ m vf2 + mgyf
Notice that if we only consider gravity as the external force then
the x and z velocities remain constant
To ½ m vyi2 + mgyi = ½ m vyf2 + mgyf
Add ½ m vxi2 + ½ m vzi2 and ½ m vxf2 + ½ m vzf2
½ m vi2 + mgyi = ½ m vf2 + mgyf
where vi2 ≡ vxi2 +vyi2 + vzi2
½ m v2 = K terms are defined to be kinetic energies
(A scalar quantity of motion)

5. What is the SI unit of energy A erg B dyne C N s D calorie E Joule
Reading Quiz
What is the SI unit of energy
A erg
B dyne
C N s
D calorie
E Joule

6. Inelastic collision in 1-D: Example 1
A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V.
What is the initial energy of the system ?
What is the final energy of the system ?
Is energy conserved? x v V
before
after

7. Inelastic collision in 1-D: Example 1
What is the momentum of the bullet with speed v ?
What is the initial energy of the system ?
What is the final energy of the system ?
Is momentum conserved (yes)?
Is energy conserved? Examine Ebefore-Eafter v
No! V x
before
after

8. Elastic vs. Inelastic Collisions
A collision is said to be inelastic when “mechanical” energy ( K +U ) is not conserved before and after the collision.
How, if no net Force then momentum will be conserved.
Kbefore  Kafter
E.g. car crashes on ice: Collisions where objects stick together
A collision is said to be elastic when both energy & momentum are conserved before and after the collision Kbefore = Kafter
Carts colliding with a perfect spring, billiard balls, etc.

9. Inelastic collision: Where did that energy go?

10. Kinetic & Potential energies
Kinetic energy, K = ½ mv2, is defined to be the large scale collective motion of one or a set of masses
Potential energy, U, is defined to be the “hidden” energy in an object which, in principle, can be converted back to kinetic energy
Mechanical energy, EMech, is defined to be the sum of U and K.

11. ½ m vyi2 + mgyi = ½ m vyf2 + mgyf
Energy
If only “conservative” forces are present, the total mechanical energy
(sum of potential, U, and kinetic energies, K) of a system is conserved
For an object in a gravitational “field”
½ m vyi2 + mgyi = ½ m vyf2 + mgyf
U ≡ mgy
K ≡ ½ mv2
Emech = K + U
Emech = K + U = constant
K and U may change, but Emech = K + U remains a fixed value.
Emech is called “mechanical energy”

12. Example of a conservative system: The simple pendulum.
Suppose we release a mass m from rest a distance h1 above its lowest possible point.
What is the maximum speed of the mass and where does this happen ?
To what height h2 does it rise on the other side ? v h1 h2 m

13. Example: The simple pendulum.
What is the maximum speed of the mass and where does this happen ?
E = K + U = constant and so K is maximum when U is a minimum. y
y=h1
y=0

14. Example: The simple pendulum.
What is the maximum speed of the mass and where does this happen ?
E = K + U = constant and so K is maximum when U is a minimum
E = mgh1 at top
E = mgh1 = ½ mv2 at bottom of the swing y
y=h1 h1
y=0 v

15. Example: The simple pendulum.
To what height h2 does it rise on the other side?
E = K + U = constant and so when U is maximum again (when K = 0) it will be at its highest point.
E = mgh1 = mgh2 or h1 = h2 y
y=h1=h2
y=0

16. Potential Energy, Energy Transfer and Path
A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless
The ball is dropped
The ball slides down a straight incline
The ball slides down a curved incline
After traveling a vertical distance h, how do the three speeds compare? h 1 3 2
(A) 1 > 2 > (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell

17. Potential Energy, Energy Transfer and Path
A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless
The ball is dropped
The ball slides down a straight incline
The ball slides down a curved incline
After traveling a vertical distance h, how do the three speeds compare? h 1 3 2
(A) 1 > 2 > (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell

18. Example The Loop-the-Loop … again
To complete the loop the loop, how high do we have to let the release the car?
Condition for completing the loop the loop: Circular motion at the top of the loop (ac = v2 / R)
Exploit the fact that E = U + K = constant ! (frictionless)
h ? R
Car has mass m
Recall that “g” is the source of the centripetal acceleration and N just goes to zero is the limiting case.
Also recall the minimum speed at the top is
Ub=mgh
U=mg2R
y=0
(A) 2R (B) 3R (C) 5/2 R (D) 23/2 R
U=0

19. Example The Loop-the-Loop … again
Use E = K + U = constant
mgh + 0 = mg 2R + ½ mv2
mgh = mg 2R + ½ mgR = 5/2 mgR
h = 5/2 R
h ? R

20. Variable force devices: Hooke’s Law Springs
Springs are everywhere, (probe microscopes, DNA, an effective interaction between atoms)
In this spring, the magnitude of the force increases as the spring is further compressed (a displacement).
Hooke’s Law,
Fs = - k Du
Ds is the amount the spring is stretched or compressed from it resting position.
Rest or equilibrium position F Du

21. Exercise Hooke’s Law 8 m 9 m What is the spring constant “k” ? 50 kg
(A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m

22. SF = 0 = Fs – mg = k Du - mg Exercise 2 Hooke’s Law Fspring mg 8 m 9 m
What is the spring constant “k” ?
SF = 0 = Fs – mg = k Du - mg
Use k = mg/Du = 500 N / 1.0 m
Fspring
50 kg
(A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m mg

23. F vs. Dx relation for a foot arch:
Force (N)
Displacement (mm)

24. F-Dx relation for a single DNA molecule

25. Measurement technique: optical tweezers

26. Force vs. Energy for a Hooke’s Law spring
F = - k (x – xequilibrium)
F = ma = m dv/dt
= m (dv/dx dx/dt)
= m dv/dx v
= mv dv/dx
So - k (x – xequilibrium) dx = mv dv
Let u = x – xeq. & du = dx  m

27. Energy for a Hooke’s Law springm
Associate ½ ku2 with the “potential energy” of the spring
Ideal Hooke’s Law springs are conservative so the mechanical energy is constant

28. 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

29. Equilibrium U U Example Spring: Fx = 0 => dU / dx = 0 for x=xeq
The spring is in equilibrium position
In general: dU / dx = 0  for ANY function establishes equilibrium U U
stable equilibrium
unstable equilibrium

30. 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

31. 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

32. 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

33. Energy (with spring & gravity)1
mass: m 2 h 3 -x
When is the child’s speed greatest?
(A) At y1 (top of jump)
(B) Between y1 & y2
(C) At y2 (child first contacts spring)
(D) Between y2 & y3
(E) At y3 (maximum spring compression)

34. 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

35. Recap
Assignment:
HW6 due Tuesday Oct. 26th
For Monday: Read Ch. 11