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Lab: AP Review Sheets AP Physics Chapter 8: Potential Energy by Brenda Chen Background/Summary This chapter takes our basic knowledge of energy a step further by introducing nonisolated and isolated systems. Nonisolated systems can have energy cross boundaries, a concept called conservation of energy. Isolated systems have a total energy that is constant. We are also introduced to power. Important Notes: 1. Energy (U = mgh, K = 1/2mv2) is ALWAYS conserved! (Remember, heat is not a form of energy) 2. A nonisolated system has other forces acting on it. An isolated system does not; all the terms on its right side will equal zero. 3. Work is added to the conservation of energy equation if the system is nonisolated. Work is a method of transferring energy to a system by applying a force to a system so that a point of application undergoes a displacement. 4. A friction force transforms kinetic energy in a system to internal energy. The two are inversely proportional. 5. Power takes into account the time interval it takes to get something done. There is instantaneous power, dE/dt, and an average power, W/Ξt. The unit for power is Watts. Common Diagrams: Pendulum LcosΞΈ Important Formulae: π! + πΎ! = π! + πΎ! ΞΈ L L - LcosΞΈ Block on Table πΈ!"!#$%!!"!#!$% β π₯πΈ!"!"#$%& + π΄π!"!!"#$"%!& = πΈ!"!#$%!!"#$% πΈ!"#!!"#$!% = πΎ + π π₯πΈ = π = !πΉΒο ππ π₯πΈ!"#$%"&' = π! π ππΈ π= ππ‘ π = πΉΒο π£ 1 π! = ππ₯ ! 2 Hanging Mass m1 m2 Strategy: 1. Identify problem as using an Energy analysis. 2. Define your system: Identify initial and final positions for all objects. 3. Select zero reference points for all potential energies. 4. If any type of friction is present, mechanical energy will not be constant β account for friction using ΞEint = Ffrictionx 5. Write an equation with all energies present, and solve for the unknown. Lab: AP Review Sheets AP Physics Problem 1 (Easy): Lena is at the Pasadena Ice Rink. Her devious friend gives her a little push so that Lena is sliding at an initial speed of 3.00 m/s. The coefficient of kinetic friction between her ice skates and the ice is .140. How far does she travel until she finally reaches a stop? Solutions: 3m/s µ = .140 Remember to think of the five steps of our strategy. In this case, we are sure that this is an energy analysis problem, and we wonβt have to assign heights because they are on the same y-axis. For that reason, we also wonβt have to include the potential energy in the equation because there are staying on the same y-axis. We will, however, have to add Einternal to the equation because they gave us a coefficient of friction. π΄πΈ!"!#!$% = π΄πΈ!"#$% πΎ! β πΈ!"# = πΎ! ! ! ππ£ ! β π! π = πΎ! 0 The final kinetic energy is zero because Lena eventually comes to a stop. πΉ! = ππΉ! = πππ 1 ππ£ ! = ππππ₯ 2 π£! 3! π₯= = 2ππ 2 . 14 9.8 π₯ = 3.27 π Lab: AP Review Sheets AP Physics Problem 2 (Medium): Hana is trying to hypnotize her sister. Her 8 cm pendulum is a attached to a mass of 10 g. If the speed of the particle is 2.5 m/s at its lowest point, at what angle is the pendulum released at? What is the tension T at the lowest point? Solutions: FT Fg 1cosΞΈ ΞΈ 1m v = 2.5 m/s 10g 1-1cosΞΈ We know that we need to use the L-LcosΞΈ to find where the mass is at its lowest point. Once we established that, we can then use our conservation of energy equation. π!" + πΎ! = π!" + πΎ! 0 0 The initial kinetic energy and final potential energy cancel out because the pendulum is released from rest, and at its lowest point, we consider the height to be zero. 1 ππ£ ! 2 1 . 010 9.8 1 β 1πππ π = (.010)(2.5)! 2 1 ! 2 (.010)(2.5) = 1 β 1πππ π . 010 9.8 β.681 = 1πππ π ππ πΏ β πΏπππ π = π = 47.1 Now we need to do a force analysis to figure out the tension. Because the pendulum travels in a circular path, we need to use centripetal force. ππ£ ! π ππ£ ! πΉ! β πΉ! = π π( 2π πΏ β πΏπππ π )! π= + ππ πΏ πΉ!"# = π= (.01) (2)(9.8)(1 β 1πππ 47.1)! + (.01)(9.8) 1 π = .161 π Lab: AP Review Sheets AP Physics Problem 3 (Hard): A marshmallow gun that requires a special type of dense marshmallow has a launching mechanism that uses a spring with unknown k. When compressed only .10 m from the barrel end, the gun will launch the heavy marshmallow 2 meters above the end of the barrel. The marshmallow, which is 14 g, sits on top of the vertically oriented gun. What is the spring constant? The equilibrium position of the spring with the heavy marshmallow resting on it? The marshmallowβs speed as it moves through this position? Solutions: 2m Marshmallow (14 g) .10m Marshmallow Gun To find the spring constant, we should first assume that there is no air friction. Then we write out the conservation of energy equation (Be careful sure to use the potential energy equation for a spring) π!!!"!#!$% + π!!!"!#!$% + πΎ! = π!!!"#$! + π!!!"#$% + πΎ! 1 ! 1 1 1 ππ₯ + ππβ + ππ£ ! = ππ₯ ! + ππβ + ππ£ ! 2 2 2 2 1 ! ππ₯ = ππβ 2 The Ug-initial, Ki, Us-final, and Ug-final cancelled out because our starting point is at h = 0, the marshmallow is starting from rest, and the final speed is also 0. π= 2ππβ 2 . 014 9.8 . 1 + 2 = = 3.29 π/π ! π₯ . 1! For finding the equilibrium position, we need to use a force analysis. The right side of the equation is zero because the forces need to be equal for the system to be in equilibrium. πΉ!"# = 0 πΉ!"#$%& β πΉ! = 0 πΉ!"#$%& = βππ = βππ₯ ππ (.014)(9.8) π₯= = = .042π π (3.29 Now to find the speed of the marshmallow as it moves through this position, we would again need to do an energy analysis. π!!!"!#!$% + π!!!"!#!$% + πΎ! = π!!!"#$% + π!!!"#$% + πΎ! 1 ! 1 1 1 ππ₯ + ππβ + ππ£ ! = ππ₯ ! + ππβ + ππ£ ! 2 2 2 2 1 ! 1 ! 1 ! ππ₯ = ππ₯ + ππβ + ππ£ 2 2 2 1 1 1 ! ! (3.29)(β.1) = (3.29) β.042 + (.014)(9.8)(2 β .12) + (.014)π£ ! 2 2 2 37.3 = π£ ! = 6.10 π/π