Lecture 13

1. Lecture 13 • Goals • Introduce concepts of Kinetic and Potential energy • Develop Energy diagrams • Relate Potential energy to the external net force • Discuss Energy Transfer and Energy Conservation • Define and introduce power (energy per time)

2. Conservative vs. Non-Conservative forces • For a spring one can perform negative work but then reverse this process and recover all of this energy. • A compressed spring has the ability to do work • For a Hooke’s law spring the work done is independent of path • The spring is said to be a conservative force • In the case of friction there is no immediate way to back transfer the energy of motion • In this case the work done can be shown to be dependent on path • Friction is said to be a non-conservative force

3. Hidden energy is Potential Energy (U) • For the compressed spring the energy is “hidden” but still has the ability to do work (i.e., allow for energy transfer) • This kind of “energy” is called “Potential Energy” • The gravitation force, if constant, has the same properties.

4. Mechanical Energy (Kinetic + Potential) U ≡ mgy K ≡ ½ mv2 Emech = K + U Emech = K + U= constant • If only “conservative” forces, then total mechanical energy (potential U plus kinetic K energy) of a systemis conserved For an object in a gravitational “field” ½ m vyi2+ mgyi = ½ m vyf2 + mgyf= constant • K and U may change, but Emech = K + Uremains a fixed value. Emech is called “mechanical energy”

5. Example of a conservative system: The simple pendulum. m h1 h2 v • 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 ?

6. 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

7. 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 = mgh1at top E = mgh1 = ½ mv2 at bottom of the swing y y=h1 h1 y=0 v

8. 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

9. Potential Energy, Energy Transfer and Path 1 3 2 h • 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? (A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell

10. Energy diagrams Emech K Energy U y • In general: Ball falling Spring/Mass system Emech K Energy U 0 0 u = x - xeq

11. Conservative Forces & Potential Energy Uf rf rf ò U = Uf - Ui = - W = -F•dr ri Ui ri • For any conservative force F we can define a potential energy function U in the following way: The work done by a conservative force is equal and opposite to the change in the potential energy function. • This can be written as: ò W=F·dr= - U

12. Conservative Forces and Potential Energy • So we can also describe work and changes in potential energy (for conservative forces) DU = - W • Recalling (if 1D) W = Fx Dx • Combining these two, DU = - Fx Dx • Letting small quantities go to infinitesimals, dU = - Fx dx • Or, Fx = -dU / dx

13. 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 stable equilibrium unstable equilibrium

14. Chapter 8 Conservation of Energy Examples of energy transfer • Work (by conservative or non-conservative forces) • Heat (thermal energy transfer) • A “system” depends on situation • Example: A can with internal non-conservative forces In general, if just one external force acting on a system

15. “Mechanical” Energy of a System…agai n • Isolated system without non-conservative forces

16. ExampleThe Loop-the-Loop … again Ub=mgh 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 Car has mass m U=mg2R h ? R • 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 thatE = U + K = constant ! (frictionless) (A) 2R (B) 3R (C) 5/2 R (D) 23/2 R y=0 U=0

17. ExampleThe 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

18. With non-conservative forces • Mechanical energy is not conserved.

19. An experiment N T T m2 fk m1 m2g m1g Two blocks are connected on the table as shown. The table has a kinetic friction coefficient of mk. The masses start at rest and m1 falls a distance d. How fast is m2 going? Mass 1 S Fy = m1ay = T – m1g Mass 2 S Fx = m2ax = -T + fk= -T + mkN S Fy = 0 = N – m2g | ay | = | ay | = a =(mkm2 - m1) / (m1 + m2) 2ad = v2 =2(mkm2 - m1) g / (m1 + m2) DK= - mkm2gd – Td + Td + m1gd = ½ m1v2+ ½ m2v2 v2 =2(mkm2 - m1) g / (m1 + m2)

20. Energy conservation for a Hooke’s Law spring m • Associate ½ kx2 with the “potential energy” of the spring • Ideal Hooke’s Law springs are conservative so the mechanical energy is constant

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

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

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

24. Energy (with spring & gravity) 1 mass: m • 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) 2 h 3 0 -x

25. Work & Power: • Two cars go up a hill, a Corvette and a ordinary Chevy Malibu. Both cars have the same mass. • Assuming identical friction, both engines do the same amount of work to get up the hill. • Are the cars essentially the same ? • NO. The Corvette can get up the hill quicker • It has a more powerful engine.

26. Work & Power: • Power is the rate at which work is done. • Average Power is, • Instantaneous Power is, • If force constant in 1D, W= F Dx = F (v0Dt + ½ aDt2) and P = W / Dt = F (v0 + aDt) 1 W = 1 J / 1s

27. Exercise Work & Power Power time Power Z3 time Power time • Top • Middle • Bottom • Starting from rest, a car drives up a hill at constant acceleration and then suddenly stops at the top. • The instantaneous power delivered by the engine during this drive looks like which of the following,

28. Exercise Work & Power • P = dW / dt and W = F d = (mmg cosq - mg sinq) d and d = ½ a t2 (constant accelation) So W = F ½ a t2  P = F a t = F v • (A) • (B) • (C) Power time Power Z3 time Power time

29. Work & Power: • Power is the rate at which work is done. Example: • A person of mass 80.0 kg walks up to 3rd floor (12.0m). If he/she climbs in 20.0 sec what is the average power used. • Pavg = F h / t = mgh / t = 80.0 x 9.80 x 12.0 / 20.0 W • P = 470. W

30. Recap • Read through Chap 9.1-9.4