1. Physics 201: Lecture 13, Pg 1 Lecture 13 l 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. Physics 201: Lecture 13, Pg 2 Conservative vs. Non-Conservative forces l For a spring one can perform negative work but then reverse this process and recover all of this energy. l A compressed spring has the ability to do work l For a Hooke’s law spring the work done is independent of path l The spring is said to be a conservative force l In the case of friction there is no immediate way to back transfer the energy of motion l In this case the work done can be shown to be dependent on path l Friction is said to be a non-conservative force

3. Physics 201: Lecture 13, Pg 3 Hidden energy is Potential Energy (U) l For the compressed spring the energy is “hidden” but still has the ability to do work (i.e., allow for energy transfer) l This kind of “energy” is called “Potential Energy” l The gravitation force, if constant, has the same properties.

4. Physics 201: Lecture 13, Pg 4 Mechanical Energy (Kinetic + Potential) l If only “conservative” forces, then total mechanical energy () of a system is conserved (potential U plus kinetic K energy) of a system is conserved For an object in a gravitational “field” l K and U may change, but E mech = K + U remains a fixed value. E mech = K + U = constant E mech is called “mechanical energy” E mech = K + U K ≡ ½ mv 2 U ≡ mgy ½ m v yi 2 + mgy i = ½ m v yf 2 + mgy f = constant

5. Physics 201: Lecture 13, Pg 5 Example of a conservative system: The simple pendulum. l Suppose we release a mass m from rest a distance h 1 above its lowest possible point.  What is the maximum speed of the mass and where does this happen ?  To what height h 2 does it rise on the other side ? v h1h1 h2h2 m

6. Physics 201: Lecture 13, Pg 6 Example: The simple pendulum. y y= 0 y=h 1  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.

7. Physics 201: Lecture 13, Pg 7 Example: The simple pendulum. v h1h1 y y=h 1 y=0  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 = mgh 1 at top E = mgh 1 = ½ mv 2 at bottom of the swing

8. Physics 201: Lecture 13, Pg 8 Example: The simple pendulum. y y=h 1 =h 2 y=0 To what height h 2 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 = mgh 1 = mgh 2 or h 1 = h 2

9. Physics 201: Lecture 13, Pg 9 Potential Energy, Energy Transfer and Path l A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless 1. The ball is dropped 2. The ball slides down a straight incline 3. 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 h 13 2

10. Physics 201: Lecture 13, Pg 10 Energy diagrams l In general: Energy K y U E mech Energy K u = x - x eq U E mech Spring/Mass system Ball falling 0 0

11. Physics 201: Lecture 13, Pg 11 Conservative Forces & Potential Energy l 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. l This can be written as: W = F ·dr = -  U   U = U f - U i = - W = - F dr  riri rfrf riri rfrf UfUf UiUi

12. Physics 201: Lecture 13, Pg 12 Conservative Forces and Potential Energy l So we can also describe work and changes in potential energy (for conservative forces)  U = - W l Recalling (if 1D) W = F x  x l Combining these two,  U = - F x  x l Letting small quantities go to infinitesimals, dU = - F x dx l Or, F x = -dU / dx

13. Physics 201: Lecture 13, Pg 13 Equilibrium l Example  Spring: F x = 0 => dU / dx = 0 for x=x eq The spring is in equilibrium position l In general: dU / dx = 0  for ANY function establishes equilibrium stable equilibrium unstable equilibrium U U

14. Physics 201: Lecture 13, Pg 14 Chapter 8 Conservation of Energy Examples of energy transfer l Work (by conservative or non-conservative forces) l Heat (thermal energy transfer) l A “system” depends on situation l Example: A can with internal non-conservative forces In general, if just one external force acting on a system

15. Physics 201: Lecture 13, Pg 15 “Mechanical” Energy of a System…agai n l Isolated system without non-conservative forces

16. Physics 201: Lecture 13, Pg 16 Example The Loop-the-Loop … again l To complete the loop the loop, how high do we have to let the release the car? l Condition for completing the loop the loop: Circular motion at the top of the loop (a c = v 2 / R) l Exploit the fact that E = U + K = constant ! (frictionless) (A) 2R (B) 3R(C) 5/2 R(D) 2 3/2 R 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 U b =mgh U=mg2R y=0 U=0

17. Physics 201: Lecture 13, Pg 17 Example The Loop-the-Loop … again l Use E = K + U = constant l mgh + 0 = mg 2R + ½ mv 2 mgh = mg 2R + ½ mgR = 5/2 mgR h = 5/2 R R h ?

18. Physics 201: Lecture 13, Pg 18 With non-conservative forces l Mechanical energy is not conserved.

19. Physics 201: Lecture 13, Pg 19 An experiment Two blocks are connected on the table as shown. The table has a kinetic friction coefficient of  k. The masses start at rest and m 1 falls a distance d. How fast is m 2 going? T m1m1 m2m2 m2gm2g N m1gm1g T fkfk Mass 1  F y = m 1 a y = T – m 1 g Mass 2  F x = m 2 a x = -T + f k = -T +  k N  F y = 0 = N – m 2 g | a y | = | a y | = a =(  k m 2 - m 1 ) / (m 1 + m 2 ) 2ad = v 2 =2 (  k m 2 - m 1 ) g / (m 1 + m 2 )  K= -  k m 2 gd – Td + Td + m 1 gd = ½ m 1 v 2 + ½ m 2 v 2 v 2 =2 (  k m 2 - m 1 ) g / (m 1 + m 2 )

20. Physics 201: Lecture 13, Pg 20 Energy conservation for a Hooke’s Law spring l Associate ½ kx 2 with the “potential energy” of the spring m l Ideal Hooke’s Law springs are conservative so the mechanical energy is constant

21. Physics 201: Lecture 13, Pg 21 Energy (with spring & gravity) l E mech = constant (only conservative forces) l At 1: y 1 = h ; v 1y = 0 At 2: y 2 = 0 ; v 2y = ? At 3: y 3 = -x ; v 3 = 0 l E m1 = U g1 + U s1 + K 1 = mgh + 0 + 0 l E m2 = U g2 + U s2 + K 2 = 0 + 0 + ½ mv 2 l E m3 = U g3 + U s3 + K 3 = -mgx + ½ kx 2 + 0 1 3 2 h 0 -x mass: m Given m, g, h & k, how much does the spring compress?

22. Physics 201: Lecture 13, Pg 22 Energy (with spring & gravity) l E mech = constant (only conservative forces) l At 1: y 1 = h ; v 1y = 0 At 2: y 2 = 0 ; v 2y = ? At 3: y 3 = -x ; v 3 = 0 l E m1 = U g1 + U s1 + K 1 = mgh + 0 + 0 l E m2 = U g2 + U s2 + K 2 = 0 + 0 + ½ mv 2 l E m3 = U g3 + U s3 + K 3 = -mgx + ½ kx 2 + 0 l Given m, g, h & k, how much does the spring compress? l E m1 = E m3 = mgh = -mgx + ½ kx 2  Solve ½ kx 2 – mgx +mgh = 0 1 3 2 h 0 -x mass: m Given m, g, h & k, how much does the spring compress?

23. Physics 201: Lecture 13, Pg 23 Energy (with spring & gravity) l E mech = constant (only conservative forces) l At 1: y 1 = h ; v 1y = 0 At 2: y 2 = 0 ; v 2y = ? At 3: y 3 = -x ; v 3 = 0 l E m1 = U g1 + U s1 + K 1 = mgh + 0 + 0 l E m2 = U g2 + U s2 + K 2 = 0 + 0 + ½ mv 2 l E m3 = U g3 + U s3 + K 3 = -mgx + ½ kx 2 + 0 l Given m, g, h & k, how much does the spring compress? l E m1 = E m3 = mgh = -mgx + ½ kx 2  Solve ½ kx 2 – mgx +mgh = 0 1 3 2 h 0 -x mass: m Given m, g, h & k, how much does the spring compress?

24. Physics 201: Lecture 13, Pg 24 Energy (with spring & gravity) l When is the child’s speed greatest? (A) At y 1 (top of jump) (B) Between y 1 & y 2 (C) At y 2 (child first contacts spring) (D) Between y 2 & y 3 (E) At y 3 (maximum spring compression) 1 3 2 h 0 -x mass: m

25. Physics 201: Lecture 13, Pg 25 Work & Power: l Two cars go up a hill, a Corvette and a ordinary Chevy Malibu. Both cars have the same mass. l Assuming identical friction, both engines do the same amount of work to get up the hill. l Are the cars essentially the same ? l NO. The Corvette can get up the hill quicker l It has a more powerful engine.

26. Physics 201: Lecture 13, Pg 26 Work & Power: l Power is the rate at which work is done. l Average Power is, l Instantaneous Power is, If force constant in 1D, W= F  x = F (v 0  t + ½ a  t 2 ) and P = W /  t = F (v 0 + a  t) 1 W = 1 J / 1s

27. Physics 201: Lecture 13, Pg 27 Exercise Work & Power A. Top B. Middle C. Bottom l Starting from rest, a car drives up a hill at constant acceleration and then suddenly stops at the top. l The instantaneous power delivered by the engine during this drive looks like which of the following, Z3 time Power time

28. Physics 201: Lecture 13, Pg 28 Exercise Work & Power P = dW / dt and W = F d = (  mg cos  mg sin  d and d = ½ a t 2 (constant accelation) So W = F ½ a t 2  P = F a t = F v l (A) l (B) l (C) Z3 time Power time

29. Physics 201: Lecture 13, Pg 29 Work & Power: l Power is the rate at which work is done. l 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. l P avg = F h / t = mgh / t = 80.0 x 9.80 x 12.0 / 20.0 W l P = 470. W Example:

30. Physics 201: Lecture 13, Pg 30 Recap l Read through Chap 9.1-9.4