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Unit 5 - Work and Energy CHAPTER 8 CONCEPTUAL PHYSICS BOOK CHAPTER 6 PHYSICS BOOK Part 2 CONSERVATION OF ENERGY AND POWER Conservative vs. Nonconservative Forces ο΅ Conservative force – total Work on a closed path is zero. (ex: gravity) Gravity- down Motion- up ο΅ -W +W Gravity- down Motion- down Nonconservative force – total Work on a closed path is NOT zero. (ex: friction) Friction - right Motion- left -W -W Energy 3 Friction – left Motion - right Conservation of Energy ο΅ Law of Conservation of Energy – Energy cannot be created or destroyed, only converted from one form to another. ο΅ This means the amount of energy when everything started is still the amount of energy in the universe today! (Just in different forms!) π¬π = π¬π Conservation of Mechanical Energy ο΅ If non-conservative forces are NOT present (or are ignored) the total Mechanical Energy initially is equal to the total Mechanical Energy final. π»π΄π¬π = π»π΄π¬π OR π²π¬π + π·π¬π = π²π¬π + π·π¬π Conceptual Example 1: Pendulum ο΅ Pendulum - Kinetic and Potential Energy ο΅ In the absence of air resistance and friction… ο΅ the ο΅ pendulum would swing forever ο΅ example of conservation of mechanical energy ο΅ Potential → Kinetic → Potential and so on… In reality, air resistance and friction cause mechanical energy loss, so the pendulum will eventually stop. Conceptual Example 2: Roller Coaster ο΅ Roller Coaster - Kinetic and Potential Energy With Non-Conservative Forces… ο΅ If non-conservative forces (such as friction or air resistance) ARE present: π΄π¬π = π΄π¬π + πΎππ ο΅ Be careful: Work done by friction is always negative! (Friction always opposes the motion) ο΅ So if friction is present, there is mechanical energy loss. (The energy is converted into heat and sound.) Conceptual Example 3: Downhill Skiing ο΅ Downhill Skiing - Kinetic and Potential Energy ο΅ This animation neglects friction and air resistance until the bottom of the hill. ο΅ Friction is provided by the unpacked snow. ο΅ Mechanical ο΅ Negative energy loss (nonconservative force) work Problem Solving Insights ο΅ ο΅ Determine if non-conservative forces are included. ο΅ If yes: MEf = ME0 + Wnc ο΅ If no: (We won’t be solving this type) MEf = ME0 Eliminate pieces that are zero before solving ο΅ Key words: starts from rest (KE0 = 0), ends on the ground (PEf = 0), etc. Example 1 ο΅ A 2.00kg rock is released from rest from a height of 20.0 m. Ignore air resistance & determine the kinetic, potential, & mechanical energy at each of the following heights: 20.0 m, 12.0m, 0m (Round g to 10 m/s2 for ease) Example 1 - Answers Height 20.0 m KE PE ME 0J Start Here 2*10*20 = 400 J 400 J 12.0 m 400-240 = 160 J 2*10*12 = 240 J 400 J 0m 400-0 = 400 J 2*10*0 = 0 J 400 J Then Use This 13 Example 2 Find the potential energy, kinetic energy, mechanical energy, velocity, and height of the skater at the various locations below. Energy max 14 Example 2 - Answers 1. β = 0, so ππΈ = 0 π½ 3. π£ = 0 at the top, so πΎπΈ = 0 π½ 1 2 ππΈ = 1920 π½ ππΈ = 1920 − 0 = 1920 π½ 1920 = 60 9.8 β so β ≈ 3.3 π πΎπΈ = 60 8 2 = 1920 π½ ππΈ = 0 + 1920 = 1920 π½ 2. β = 1, so ππΈ = 60 9.8 1 = 588 π½ ππΈ = 1920 π½ πΎπΈ = 1920 − 588 = 1332 π½ 1 π 2 1332 = (60)π£ so π£ ≈ 6.7 2 π Power ο΅ ο΅ Power: Rate of doing work. The work done per unit time. Equation: π = ο΅P ο΅ π π‘ is power (Watts, ft lb/s , ft lb/min) Horsepower: another unit for measuring power. ο΅1 horsepower = 746 Watts (or 1 horsepower = 550 ftlb/s) Power Example #1 ο΅ A weight lifter lifts a 75 kg weight from the ground to a height of 2.0 m. He performs this feat in 1.5 seconds. Find the weight lifter’s average power in A) Watts and B)Horsepower. ο΅ π = πΉπ = 75 9.8 2 = 1470 π½ ο΅ ο΅ A. π = B. 980 746 π π‘ = 1470 1.5 ≈ 1.3 π»π = 980 πππ‘π‘π Power Example #2 ο΅ ο΅ A runner sprints 100 m in 25 seconds. Her average power during this run is 800 Watts. Find the force that the runner exerts during the run. π= πΉπ π‘ πΉ∗100 25 ο΅ 800 = ο΅ 20,000 = 100πΉ ο΅ πΉ = 200 π Power Example #3 ο΅ A car accelerates from rest to 20.0 m/s is 5.6 seconds along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if the weight of the car is 9,000 N 9000 ο΅ π= ≈ 918 ππ 9.8 1 1 1 2 2 ο΅ π = ππ£ − ππ£0 = (918) 20 2 2 2 2 π 184000 ο΅ π= = = 32,800 πππ‘π‘π π‘ 5.6 ≈ 184,000 Power Example #4 ο΅ Bob pushes a box across a horizontal surface at a constant speed of 1 m/s. If the box has a mass of 30 kg, find the power Bob supplies if the coefficient of kinetic friction is 0.3. ο΅ Since a=0, the pushing force must be equal to the kinetic friction. ο΅ π = ππ = 0.3 30 9.8 = 88.2 π ο΅ 1 m/s implies that after 1 second the distance is 1 meter, so… ο΅ π= πΉπ π‘ = 88.2 (1) (1) = 88.2 πππ‘π‘π
