Physics 101: Lecture 13, Pg 1 Physics 101: Lecture 13 l Quick Review of Last Time, Example Problems l Power, Work done by a variable force l Reminders: çExam I, Tuesday, September 30 th at 5 PM çSee PHY101 Web page for room assignments çPlease do not forget to bring your UB ID card ! l Chapter 6 : Work and Energy

Physics 101: Lecture 13, Pg 2 Work done by a constant Force   KE = W net W = F s = |F| |s| cos  = F s s |F| : magnitude of force |s| = s : magnitude of displacement F s = magnitude of force in direction of displacement : F s = |F| cos   : angle between displacement and force vectors Kinetic energy : KE= 1/2 m v 2 Work-Kinetic Energy Theorem: F s

Physics 101: Lecture 13, Pg 3 Work Done by Gravity l Example 1: Drop ball Y i = h Y f = h f W g = (mg)(S)cos  S = h 0 -h f W g = mg(h 0 -h f ) cos(0 0 ) = mg(h 0 -h f ) = PE initial – PE final mg S y x Y i = h 0 mg S y x

Physics 101: Lecture 13, Pg 4 Work Done by Gravity Example 2: Toss ball up W g = (mg)(S)cos  S = h 0 -h f W g = mg(h 0 -h f )cos(180 0 ) = =-mg(h 0 -h f ) = PE initial – PE final Y i = h 0 Y f = h f mg S y x

Physics 101: Lecture 13, Pg 5 Work Done by Gravity Example 3: Slide block down incline W g = (mg)(S)cos  S = h/cos  W g = mg(h/cos  )cos  W g = mgh with h= h 0 -h f h  mg S  Work done by gravity is independent of path taken between h 0 and h f => The gravitational force is a conservative force. h0h0 h f

Physics 101: Lecture 13, Pg 6 Concept Question Imagine that you are comparing three different ways of having a ball move down through the same height. In which case does the ball reach the bottom with the highest speed? 1. Dropping 2. Slide on ramp (no friction) 3. Swinging down 4. All the same In all three experiments, the balls fall from the same height and therefore the same amount of their gravitational potential energy is converted to kinetic energy. If their kinetic energies are all the same, and their masses are the same, the balls must all have the same speed at the end correct

Physics 101: Lecture 13, Pg 7 Conservation of Mechanical Energy l Total mechanical energy of an object remains constant provided the net work done by non-conservative forces is zero: E tot = E kin + E pot = constant or E kin,f +E pot,f = E kin,0 +E pot,0 Otherwise, in the presence of net work done by non-conservative forces (e.g. friction): W nc = E kin,f – E kin,0 + E pot,f -E pot,i

Physics 101: Lecture 13, Pg 8 Example Problem Suppose the initial kinetic and potential energies of a system are 75J and 250J respectively, and that the final kinetic and potential energies of the same system are 300J and -25J respectively. How much work was done on the system by non-conservative forces? 1. 0J 2. 50J J J J correct Work done by non-conservative forces equals the difference between final and initial kinetic energies plus the difference between the final and initial gravitational potential energies. W = (300-75) + ((-25) - 250) = = -50J.

Physics 101: Lecture 13, Pg 9 Power l Average power is the average rate at which a net force does work: P av = W net / t SI unit: [P] = J/s = watt (W) Or P av = F net s /t = F net v av

Physics 101: Lecture 13, Pg 10 Work done by a Variable Force l The magnitude of the force now depends on the displacement: F s (s) Then the work done by this force is equal to the area under the graph of F s versus s, which can be approximated as follows: W =  W i  F s (s i )  s = (F s (s 1 )+F s (s 2 )+…)  s

Physics 101: Lecture 13, Pg 11 Example Problems: l [C&J Chapter 6, problem 10] P = 256 N => W net = 0 and F net,x = 0 l [C&J Chapter 6, problem 78] Energy conservation (F N =0 => W nc =0 ): E tot,f = E tot,i  m g r cos  + ½ m v f 2 = m g r (1) Centripetal force is provided only by gravitational force (FN=0) : - m g cos  = -m v f 2 /r (2) => Solve Eq.(2) for v f 2, plug in Eq.(1) and solve for  : cos  = 2/3 =>  = 48 degrees