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Mechanical Energy Forms: Kinetic and Potential Energy Lesson 9 Announcements HW #8 due today HW #9 due tomorrow Lunch Bunch – Waves #2 this week Waves HW #1 due Tuesday HW Quiz TUESDAY Optics Exam repair Friday 7am, 4th, 5th, 3pm AP Physics B Standards Mechanical Energy forms: Kinetic and Potential Energy LESSON 9: I.C.1. Work and the Work-Energy Theorem b)Students should understand and be able to apply the work-energy theorem, so they can: (1) Calculate the change in kinetic energy or speed that results from performing a specified amount of work on an object. (2) Calculate the work performed by the net force, or by each of the forces that make up the net force, on an object that undergoes a specified change in speed or kinetic energy. (3) Apply the theorem to determine the change in an object’s kinetic energy and speed that results from the application of specified forces, or to determine the force that is required in order to bring an object to rest in a specified distance. Lesson Objectives Students will be able to 1. 2. distinguish between conservative and non-conservative forces apply the work-energy theorem to calculate changes in kinetic energy due to net external forces. Force types Forces acting on a system can be divided into two types according to how they affect potential energy. Conservative forces can be related to potential energy changes. Non-conservative forces cannot be related to potential energy changes. So, how exactly do we distinguish between these two types of forces? Conservative forces Work is path independent. Work along a closed path is zero. If the starting and ending points are the same, no work is done by the force. Work changes potential energy. Examples: Work can be calculated from the starting and ending points only. The actual path is ignored in calculations. Weight Spring force Conservation of mechanical energy holds! More on paths and conservative forces. Q: Assume a conservative force moves an object along the various paths. Which two works are equal? A: W2 = W3 (path independence) Q: Which two works, when added together, give a sum of zero? A: W1 + W2 = 0 or W1 + W3 = 0 (work along a closed path is zero) Non-conservative forces Work is path dependent. Work along a closed path is NOT zero. Work changes mechanical energy. Examples: Knowing the starting and ending points is not sufficient to calculate the work. Friction Drag (air resistance) Conservation of mechanical energy does not hold! The Work-Energy Theorem The net work due to all forces equals the change in the kinetic energy of a system. Wnet = K Wnet: work due to all forces acting on an object K: change in kinetic energy (Kf – Ki) Sample Problem: A 15-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s. a) What would the speed of the acorn have been if there had been no air resistance? b) Did air resistance do positive, negative or zero work on the acorn? Why? Sample Problem: A 15-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s. c) How much work was done by air resistance? d) What was the average force of air resistance? Conservative forces and Potential energy Potential energy is energy of position or configuration. Sometimes called “stored” energy For gravity: Ug = mgh m: mass g: acceleration due to gravity h: height above the “zero” point For springs: Us = ½ k x2 k: spring force constant x: displacement from equilibrium position Conservative forces and Potential energy Conservative forces change the potential energy of a system. Wc U If a conservative force does positive work on a system, potential energy is lost. If a conservative force does negative work, potential energy is gained. Sample Problem: A vertical spring stores 0.962 J in spring potential energy when a 3.0 kg mass is suspended from it. a) By what factor does the spring potential energy change if the mass attached to the spring is doubled? Us = ½ k x2, which seems mass independent. But mass changes the displacement! Fs – mg = 0 k x = mg x = mg/k If m is doubled, x is doubled, and since Us is proportional to x2, the energy is quadrupled. b) Verify your answer to part (a) by calculating the spring potential energy when a 6.0 kg mass is attached to the spring. 1) Solve for k Us = ½ k x2 Us = ½ k (mg/k)2 0.962 J = ½ k (3.0 kg*9.8/k)2 2*0.962 /(3*9.8)2 = k/k2 k = 450 N/m 2) Solve for Us Us = ½ k x2 Us = ½ k (mg/k)2 Us = ½ (450 N/m)[(6.0 kg * 9.8 m/s2 / 450 N/m)2 Us = 3.842 J 3) Compare to initial Us 3.842/4 = 0.960, which is close enough to 0.962 Sample Problem: A candy bar called the Mountain Bar has a calorie content of 210.0 Cal = 210 kCal, which is equivalent to 8.791 x 105 J. If an 82.0 kg mountain climber eats a Mountain Bar and magically converts it all to potential energy, what gain of altitude would be possible? Sample Problem: As an Acapulco cliff diver drops to the water from a height of 40.0 m, his gravitational potential energy decreases by 25,000 J. How much does the diver weigh?