2. Summary Lecture 7 7.1-7.6Work and Kinetic energy 8.2Potential energy 8.3Conservative Forces and Potential energy 8.5Conservation of Mech. Energy 8.6Potential-energy curves 8.8Conservation of Energy Systems of Particles 9.2Centre of mass 7.1-7.6Work and Kinetic energy 8.2Potential energy 8.3Conservative Forces and Potential energy 8.5Conservation of Mech. Energy 8.6Potential-energy curves 8.8Conservation of Energy Systems of Particles 9.2Centre of mass Problems: Chap. 8 5, 8, 22, 29, 36, 71, 51 Chap. 9 1, 6, 82 Problems: Chap. 8 5, 8, 22, 29, 36, 71, 51 Chap. 9 1, 6, 82 Thursdays 12 – 2 pm PPP “Extension” lecture. Room 211 podium level Turn up any time Thursdays 12 – 2 pm PPP “Extension” lecture. Room 211 podium level Turn up any time

3. Outline Lecture 7 Work and Kinetic energy Work done by a net force results in kinetic energy Some examples: gravity, spring, friction Potential energy Work done by some (conservative) forces can be retrieved. This leads to the principle that energy is conserved Conservation of Energy Potential-energy curves The dependence of the conservative force on position is related to the position dependence of the PE F(x) = -d(U)/dx

4. Kinetic Energy Work-Kinetic Energy Theorem Change in KE  work done by all forces  K   w

5. xixi xfxf = 1/2mv f 2 – 1/2mv i 2 = K f - K i  K =  K Work done by net force = change in KE Work-Kinetic Energy Theorem FF x Vector sum of all forces acting on the body

6. mg F h Lift mass m with constant velocity Work done by me (take down as +ve) = F.(-h) = -mg(-h) = mgh Work done by gravity = mg.(-h) = -mgh ________ Total work by ALL forces (  W) = 0 What happens if I let go? =K=K Gravitation and work Work done by ALL forces = change in KE  W =  K

7. Compressing a spring Compress a spring by an amount x Work done by me  Fdx =  kxdx = 1/2kx 2 Work done by spring  -kxdx =-1/2kx 2 Total work done (  W) = 0 =K=K What happens if I let go? x F-kx

8. F f d Work done by me = F.d Work done by friction = -f.d = -F.d Total work done = 0 What happens if I let go?NOTHING!! Gravity and spring forces are Conservative Friction is NOT!! Moving a block against friction at constant velocity

9. A force is conservative if the work it does on a particle that moves through a round trip is zero: otherwise the force is non-conservative A force is conservative if the work done by it on a particle that moves between two points is the same for all paths connecting these points: otherwise the force is non-conservative. Conservative Forces

10. A force is conservative if the work it does on a particle that moves through a round trip is zero; otherwise the force is non-conservative Conservative Forces work done by gravity for round trip: On way up: work done by gravity = -mgh On way down: work done by gravity = mgh Total work done = 0 Sometimes written as h -g Consider throwing a mass up a height h

11. Conservative Forces -g Each step height=  h = -mg(  h 1 +  h 2 +  h 3 +……) = -mgh Same as direct path (-mgh) Work done by gravity w = -mg  h 1 + -mg  h 2 +-mg  h 3 +… h A force is conservative if the work done by it on a particle that moves between two points is the same for all paths connecting these points: otherwise the force is non-conservative.

13.  U = -  w Lift mass m with constant velocity Work done by gravity = mg.(-h) = -mgh Potential Energy The change in potential energy is equal to minus the work done BY the conservative force ON the body. Therefore change in PE is  U = -  w h mg  U grav = +mgh

14. Work done by spring is  w =  -kx dx = - ½ kx 2 Potential Energy The change in potential energy is equal to minus the work done BY the conservative force ON the body. Therefore the change in PE is  U = -  w Compress a spring by an amount x F-kx x  U spring = + ½ kx 2

15. Potential Energy The change in potential energy is equal to minus the work done BY the conservative force ON the body.  U = -  w but recall that  w =  K so that  U = -  K or  U +  K = 0 Any increase in PE results from a decrease in KE decrease n increase

16.  U +  K = 0 Let’s check this for a body of mass m moving under gravity. xixi xfxf h mg  w =  K = K f - K i  K = ½ mv f 2 – ½ mv i 2 For motion under gravity you know v 2 = u 2 + 2as  v f 2 = v i 2 - 2gh mult by ½ m  ½ m v f 2 = ½ mv i 2 -mgh  K = ½ mv i 2 -mgh – ½ mv i 2  K = -mgh +ve =  w = -  U so  U +  K = 0 vfvf vivi

17.  U +  K = 0 In a system of conservative forces, any change in Potential energy is compensated for by an inverse change in Kinetic energy U + K = E In a system of conservative forces, the mechanical energy remains constant

18. Potential-energy diagrams  w  = -  U The force is the negative gradient of the PE curve If we know how the PE varies with position, we can find the conservative force as a function of position In the limit thus = F.  x

19. Energy x U= ½ kx 2 PE of a spring F = -kx (spring force) here U = ½ kx 2

20. Energy x Potential energy U = ½ kx 2 U = ½ kA 2 x=A KE PE At any position x PE + KE = E U + K = E K = E - U = ½ kA 2 – ½ kx 2 = ½ k(A 2 -x 2 ) x’ Total mech. energy E= ½ kA 2

21. EtEt K U F net =-dU/dt F net = mg – R R = mg - F net Roller Coaster

22. EtEt K U F net =-dU/dx F net = mg – R R = mg - F net mg R

23. Conservation of Energy We said: when conservative forces act on a body  U +  K = 0  U + K = E (const) This would mean that a pendulum would swing for ever. In the real world this does not happen.

24. Conservation of Energy When non-conservative forces are involved, energy can appear in forms other than PE and KE (e.g. heat from friction)  U +  K +  U int = 0  K i + U i = K f + U f + U int Energy converted to other forms Energy may be transformed from one kind to another in an isolated system, but it cannot be created or destroyed. The total energy of the system always remains constant.

25. h f mg upward Stone thrown into air, with air resistance. How high does it go? E i = E f + E loss K i + U i = K f + U f + E loss ½mv o 2 + 0 = 0 + mgh + fh ½mv o 2 = h(mg + f) v0v0

26. h f mg downward Stone thrown into air, with air resistance. What is the final velocity ? E’ i = E’ f + E’ loss K’ i + U’ i = K’ f + U’ f + E’ loss 0 + mgh = ½mv f 2 + 0 + fh mg = ½mv f 2 + f ½mv f 2 = mg - f

29. Centre of Mass (1D) 0 x1x1 x2x2 x cm m1m1 M m2m2 M = m 1 + m 2 M x cm = m 1 x 1 + m 2 x 2 In general