1. Aim: How do we explain conservation of energy in systems under the influence of non-conservative forces?

2. Do Now:
The energies indicated on your vertical axis are evenly space; that is E3 – E2 = E2 – E1 . The energy E1 is equal to U(x1 ) and the energy E3 is equal to U(x3 ).
a) Determine the numerical values of x1 and x3 .
b) Describe the motion of the particle if its total energy is E2 .
c) What is the particle’s speed at x = x1 if its total energy E =58 J?
d) The particle is released from rest at x =1/2x. Find its speed as it passes through x = x1

3. Recall Extended Definition of Work and Energy
Wtot = Wc + WNC + Wext

4. When friction is present..
Energy is not conserved. Ei + Wkinetic friction =Ef

5. Example 1
A child of mass 25 kg slides down a slide of height 3 m. When she reaches the bottom of the slide her speed is 5 m/s. How much internal energy was produced?
Mgh-Wf=1/2 Mv2
25(10)(3)-Wf=1/2(25)(5)2
Wf = J
So J of internal energy are produced

6. Example 2
The particle shown below has a mass of 0.1 kg is released from rest at A and the surface of the bowl is rough. The speed of the particle at B is 1.50 m/s. Assume the radius of the bowl is 0.2m.
What is its kinetic energy at B?
KE=1/2mv KE=1/2(.1) (1.5)2 = J
How much mechanical energy is transformed due to friction as the particle moves from A to B?
Ui -Wf = KE
Mgh – Wf =.1125 J
.1(10)(.2)-Wf = Wf = J so J are lost due to friction
Is it possible to determine the coefficient of friction from these results in any simple manner? No

7. Problem 2

8. Example 3
A 10 kg block is released from point A. The track is frictionless except for the portion between points B and C, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2,250 N/m, and compresses the spring m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between B and C. (Solution on next page)

9. Problem 3 MgH – Wf =1/2kx2 10(10)(3) – Wf =1/2(2250)(.3)2
Wf = J
So Wf =Ff d
198.75= Ff (6)
Ff = N And Ff = µFN =µmg
33.125=µ(100)
µ=0.33

10. Example 4
A sled of mass m is given a kick on a frozen pond. The kick imparts to it an initial speed of v m/s. The coefficient of kinetic friction between the sled and ice is µ. Use energy considerations to find the distance (d) the sled moves before it stops.
½ mv2 = Wf
1/2mv2 = Ff d
1/2mv2 = µFN d
1/2mv2 =µmgd
d= v2 /(2µg)