1. Chapter 7 Potential energy and conservation of energy In this chapter we will explore one of the most fundamental concepts in physics, energy, and one of the most important principles, conservation of energy. When a conservative force does work, this work is stored in the form of potential energy (symbol: U) and can be retrieved. This is not the case with non- conservative forces. (7-1)

2. A B 11 22 O x y Potential Energy U U is defined for conservative forces only Point A (x A, y A ) Point B (x B, y B ) W AB = work done by a conservative force F from A to B W AB does not depend on the path taken. This can be expressed mathematically as follows: W AB = W(A,B) Also it can be written as: W AB = U(A) - U(B) U is the potential energy. Units: J (7-2)

3. ... xoxo x1x1 m F OAB x-axispath We assume that F depends on position i.e. F = F(x) Point A at x o is called the “reference point”. The choice of x o and U(x o ) does not affect the result of our calculations. Thus we will choose x o and U(x o ) for maximum convenience (7-3)

4. .. O A y mg dy h path m floor Example: Potential energy of the gravitational force Reference point: Point O at y o = 0 Point A at y 1 = h We choose: U(y 0 ) = 0 Thus we get : (7-4)

5. ... xoxo x m F(x) ORP x-axispath Summary If we know F(x) along the x-axis we can determine U at any point P using the following equation: The equation above gives U(x) for every point P on the x-axis with coordinate x. F(x)  U(x) (7-5)

6. What about the reverse problem? i.e. if we know U(x) can we get the force F(x)? Our starting point is the definition of U: g(x) x x’ a O Theorem from calculus: In our case U plays the role of function f and -F plays the role of function g (7-6)

7. ... xixi xfxf m F OAB x-axispath vivi vfvf Conservation of mechanical energy (7-7)

8. ... xixi xfxf m F OAB x-axispath vivi vfvf (7-8)

9. .. B A y mg h m floor U = 0 vovo Example: An object of mass m is shot straight up with an initial velocity v o. Determine the maximum height h of the object The only force acting on the object is the gravitational force. The gravitational force is conservative. (7-9)

10. Potential energy of a spring (spring constant k) (7-10)

11. Example (7-3) page 178 In the Atwood machine shown in the figure m 1 = 1.37 kg, m 2 = 1.51 kg. The system is released from rest with h 2 = 0.84 m. Find the speed v of m 2 just before it hits the floor a “before”b “after” (7-11)

12. .. AB xAxA xBxB E U x motion allowed Energy diagram An energy diagram is a plot of the potential energy U(x) versus x The blue horizontal line is the total energy E = U + K (7-12)

13. ... A B C xAxA xBxB xCxC x f O Maxima and minima of a function f(x) The function f(x) plotted in the figure has one minimum at point B and two maxima at points A and C (7-13)

14. ... A B C xAxA xBxB xCxC x U O Equilibrium Consider an object that moves along the x-axis under the action of a force whose potential energy U is plotted in the figure. (7-14)

15. (7-15) x xAxA xBxB xCxC.. A B U C FCFC FBFB. F A = 0

16. .. A B U x C FCFC FBFB. xAxA xBxB xCxC (7-16)

17. Summary 1. Motion is allowed for parts of the x-axis: 2. At turning points (points A and B) U = E At the turning points v = 0 and K = 0 3. At any point on the plot: 4. Minima of the energy diagram correspond to positions of stable equilibrium. Maxima of the energy diagram correspond to positions of unstable equilibrium (7-17) E = U+K

18. Motion in two dimensions (7-18)

19. The reverse problem: If we know U(x,y,z) can we determine F ? The recipe: (7-19)

20. We will encounter problems that fall in the following two categories: 1. All forces acting on the object(s) under study are conservative. In this case we use the expression: or simply: 2. Some forces acting on the object(s) under study are conservative and some are non-conservative. F net = F c + F nc Here F c is the vector sum of all conservative forces, and F nc is the vector sum of all non-conservative forces. In this case we use the equation: Where W nc is the work done by F nc (7-20)

21. Example (7-6) page 188 A ball of mass m = 10 kg is attached to a wire of length L = 5 m that can swing freely from a support. The ball is pulled aside so that the wire makes an angle  1 = 31  from the vertical. After 10 swings the maximum angle that the ball reaches is  2 = 25  Calculate the work W f done by air resistance (a non-conservative force) on the ball during these 10 swings (7-21)

22. Calculation of the potential energy U for the ball U = mgh h = OA - OB OA = L From triangle OBC we have; OB = OCcos  OC = L Thus OB = Lcos  and h = L(1 - cos  )  U = mgL(1 - cos  ) (7-22) O B A C h m

23.  E = W f  E = E 2 - E 1 E 1 = mgL(1 - cos  1 ) E 2 = mgL(1 - cos  2 )   E = mgL(cos  1 - cos  2 ) W f = mgL(cos  1 - cos  2 ) = 10  9.8  5  [cos(31  ) -cos(25  )] W f = -24 J (7-23) 1. “Before”2. “After” 11 22