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Transcript

Physics Unit 3 This Slideshow was developed to accompany the textbook OpenStax Physics Available for free at https://openstaxcollege.org/textbooks/college-physics By OpenStax College and Rice University 2013 edition Some examples and diagrams are taken from the textbook. Slides created by Richard Wright, Andrews Academy [email protected] Which of the following is NOT work? Pushing a Stalled Car Pulling a Wagon Climbing stairs Falling Down Carrying a Heavy Backpack Down the Hall Work The amount of force used to move something a distance Want the force in the direction of the distance 𝑊 =𝐹⋅𝑑 𝑊 = 𝐹𝑑 cos 𝜃 Unit: 𝑁 ⋅ 𝑚 = 𝐽 (Joule) Watch Eureka! 08 Marcy pulls a backpack on wheels down the 100-m hall. The 60-N force is applied at an angle of 30° above the horizontal. How much work is done by Marcy? W = 5200 J Mark is carrying books (200 N) down the 100-m hall. How much work is Mark doing on the books? W=0J The force is vertical displacement is horizontal. You carry some books (200 N) while walking down stairs height 2 m and length 3 m. How much work do you do? W = -400 J A suitcase is hanging straight down from your hand as you ride an escalator. Your hand exerts a force on the suitcase, and this force does work. Which one of the following is correct? The W is negative when you ride up and positive when you ride down The W is positive when you ride up and negative when you ride down The W is positive The W is negative Do work means W = Fd F = ma So work by a net force gives an object some acceleration Acceleration means the velocity changes 𝐹 = 𝑚𝑎 𝐹𝑑 = 𝑚𝑎𝑑 𝑣𝑓2 = 𝑣02 + 2𝑎𝑑 solve for ad 1 𝑎𝑑 = 2 𝑣𝑓2 − 𝑣02 1 𝐹𝑑 = 𝑚 2 𝑣𝑓2 − 𝑣02 1 1 𝑊 = 2 𝑚𝑣𝑓2 − 2 𝑚𝑣02 Energy is the ability to do work Kinetic Energy - Energy due to motion If something in motion hits an object, it will move it some distance 1 2 𝐾𝐸 = 𝑚𝑣 2 Scalar Unit is joule (J) Watch Eureka! 09 Work Energy Theorem Work of Net external force = change in kinetic energy 1 1 2 𝑊 = 𝑚𝑣𝑓 − 𝑚𝑣02 2 2 𝑊 = 𝐾𝐸𝑓 − 𝐾𝐸0 A 0.075-kg arrow is fired horizontally. The bowstring exerts a force on the arrow over a distance of 0.90 m. The arrow leaves the bow at 40 m/s. What average force does the bow apply to arrow? Do lots of work. 5) 3.14 × 103 𝐽 7P1-3, 5-6, 9-13 6) 1.30 × 103 𝐽 Read 7.3, 7.4 9) 9.34 × 106 𝐽, 2.33 × 109 𝐽, 250 1 7CQ7-12 10) 1.47 11) 1.1 × 1010 𝐽 Answers 1) 3 𝐽, 7.17 × 10−4 2) 1.84 × 103 𝐽 𝑚 𝑠 𝑘𝑐𝑎𝑙 3) 5.92 × 105 𝐽, −5.88 × 105 𝐽, 0 𝐽 12) −2470 𝑁, −1.48 × 105 𝑁 13) −2.8 × 103 𝑁 Potential energy Energy due to position 𝑊 = 𝐹𝑑 Gravity 𝑊𝑔𝑟𝑎𝑣𝑖𝑡𝑦 = 𝑚𝑔ℎ 𝑃𝐸 = 𝑚𝑔ℎ Since the force of gravity is down We only worry about the vertical distance Potential Energy is not absolute It is a difference The path the object takes doesn’t matter, just the vertical distance h is measured from any chosen point. Just be consistent Watch Eureka! 10 Conservative Forces A force where the work it does is independent of the path Only thing that matters is starting and stopping point Examples of conservative forces Examples of Nonconservative forces Gravitational Force Friction Elastic Spring Force Air resistance Electric Force Tension Normal force Propulsion force of things like rocket engine Each of these forces depends on the path Potential energy can be converted into Kinetic energy and back Think of an object thrown up Bottom 0 PE, high KE Top high PE, 0 KE Δ𝐾𝐸 = −Δ𝑃𝐸 𝐾𝐸𝑓 − 𝐾𝐸0 = −(𝑃𝐸𝑓 − 𝑃𝐸0 ) Rearrange Conservation of Mechanical Energy 𝐾𝐸𝑓 + 𝑃𝐸𝑓 = 𝐾𝐸0 + 𝑃𝐸0 Only if there are only conservative forces If there is no work done by nonconservative forces Total mechanical energy is constant KE0 + PE0 = KEf + PEf Other potential energy Spring Potential Energy 1 2 𝑃𝐸𝑆 = 𝑘𝑥 2 A toy gun uses a spring to shoot plastic balls 𝑚 = 50 𝑔 . The spring is compressed by 3.0 cm. Let 𝑘 = 2.22 × 105 𝑁/𝑚. (a) Of course, you have to do some work on the gun to arm it. How much work do you have to do? (b) Suppose you fire the gun horizontally. How fast does the ball leave the gun? (c) Now suppose you fire the gun straight upward. How high does the ball go? You have great potential… 18) 1.8 J, 8.6 J 7P16-18, 20-23 20) 0.687 m/s Read 7.5, 7.6 21) 26.2 m/s, 5.35 s, 26.3 m/s, 4.86 s 7CQ13-16 22) 7.81 × 105 𝑁/𝑚 23) 0.459 m Answers 1 16) 1.96 × 1016 𝐽, 𝑎𝑏𝑜𝑢𝑡 2 17) 3 × 1012 𝐽, 2 × 105 times as much Often both conservative and nonconservative forces act on an object at once. We can write Work done by net external force as 𝑊 = 𝑊𝑐 + 𝑊𝑛𝑐 𝑊𝑐 = −Δ𝑃𝐸, 𝑊 = Δ𝐾𝐸 𝑊𝑛𝑐 = Δ𝐾𝐸 + Δ𝑃𝐸 𝐾𝐸0 + 𝑃𝐸0 + 𝑊𝑛𝑐 = 𝐾𝐸𝑓 + 𝑃𝐸𝑓 𝐸0 + 𝑊𝑛𝑐 = 𝐸𝑓 Law of Conservation of Energy The total energy is constant in any process. It may change form or be transferred from one system to another, but the total remains the same Energy is transformed from one form to another Box sliding down incline PE transformed to KE KE transformed to Heat and Sound Engine Chemical to KE and Heat Efficiency Useful energy output is always less than energy input Some energy lost to heat, etc. useful energy or work output 𝑊𝑜𝑢𝑡 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝐸𝑓𝑓 = = total energy input 𝐸𝑖𝑛 A rocket starts on the ground at rest. Its final speed is 500 m/s and height is 5000 m. If the mass of the rocket stays approximately 200 kg. Find the work done by the rocket engine. W = 3.48 x 107 J A 1500-kg car’s brakes failed and it coasts down a hill from rest. The hill is 10 m high and the car has a speed of 12 m/s at the bottom of the hill. How much work did friction do on the car? Wf = -39000 J Captain Proton’s rocket pack provides 800,000 J of work to propel him from resting on his ship which is near the earth to 50 m above it. Captain Proton’s mass is 90 kg. What is his final velocity? V = 130 m/s Energy is not to be conserved while you do this homework 7P24-29 Read 7.7 7CQ18-20 Answers 24) 9.46 m/s 25) 47.6 𝑚, 1.89 × 105 𝐽, 375 𝑁 26) 4 × 104 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 27) 24.8 𝑚 𝑠 28) 1 × 104 𝑏𝑜𝑚𝑏𝑠 29) 2.5 × 107 𝑦𝑒𝑎𝑟𝑠 Two cars with the same mass do the same amount of work to get to 100 km/h. Which car is better Takes 8.0 s Takes 6.2 s Sometimes the time taken to do the work is important Rate that work is done 𝑊 𝑃= 𝑡 Unit: joule/s = watt (W) Since work changes the amount of energy in an object Power is the rate that energy is changing A 500 kg car accelerates from 0 to 100 km/h in 6.2 s on a level road. Find the average power of the car. P = 31100 W 41.73 horsepower Electrical Energy Often measured in kWh because Pt = W If it costs $0.10 per kWh, how much will it cost to run a 1000 W microwave for 2 minutes? Power through these problems in no time. 33) $2.37 7P31-40 34) $149 Read 7.8, 7.9 35) 208 W, 6.57 × 109 J 7CQ21-24 36) 208 W, 141 s 37) 643 hp Answers 38) 3.20 s, 4.04 s 31) 1 × 108 , 5 times, yes 39) 7.81 × 104 W, $7.03/h, $0.00195/s 32) 40 people, 8 million people 40) 9.46 × 107 J, 2.54 y Human bodies (all living bodies) convert energy Rate of food energy use is metabolic rate Basal metabolic rate (BMR) Total energy conversion at rest Highest: liver and spleen See table 7.4 Table 7.5 shows energy consumed for various activites Energy is required to do work World wide, the most common source of energy is oil USA has 4.5% of world population, but uses 24% of world’s oil World energy consumption continues to increase quickly Growing economies in China and India Fossil Fuels are very polluting Many countries trying to develop renewable energy like wind and solar Generally, higher energy use per capita = better standard of living You have the power to change to world, but will you work to do it? 7P45-49, 51, 56, 63, 64 Read 8.1, 8.2 48) 31 g 51) 24 W, 24.5 min, 1.14 × 106 J 56) 144 J, 288 W Answers 47) 0.100, 251 kcal 49) 3.8 × 103 kcal 8CQ1-5 45) 1.17 × 103 46) 641 W, 0.860 hp 𝑊, 1.56 hp, no 63) -21.0 N, 7.50 m, 5.77° or 84.23°, 1.50 m 64) 1.61 × 104 N, 3.22 × 105 J, 5.66 m/s, 4.00 kJ Often the force acting on an object is not constant. Baseball or Tennis ball being hit Times of force often short Force can be huge To hit a ball well Both size of force and time of contact are important Bring both these together in concept of impulse Impulse Ft Unit: Ns Is a vector Object responds to amount of impulse Large impulse Large response higher vf Large mass less velocity Both mass and velocity play role in how responds to impulse Linear Momentum p = mv Unit: kg m/s Is a vector Is important when talking about collisions 𝐹 = 𝑚𝑎 𝑣𝑓 − 𝑣0 𝑎= Δ𝑡 𝑣𝑓 − 𝑣0 𝑚𝑣𝑓 − 𝑚𝑣0 𝐹=𝑚 = Δ𝑡 Δ𝑡 𝐹Δ𝑡 = 𝑚𝑣𝑓 − 𝑚𝑣0 Impulse = Change in Momentum Hard to measure force during contact Find change in momentum Use impulse-Momentum Theorem and time of contact to find average force of contact Watch NASCAR Crash A baseball (m = 0.14 kg) with initial velocity of -40 m/s (90 mph) is hit. It leaves the bat with a velocity of 60 m/s after 0.001 s. What is the impulse and average net force applied to the ball by the bat? Impulse = 14 Ns F = 14000 N A raindrop (m = .001 kg) hits a roof of a car at -15 m/s. After it hits, it spatters so the effective final velocity is 0. The time of impact is .01 s. What is the average force? F = 1.5 N What if it is ice so that it bounces off at 10 m/s? F = 2.5 N Watch Offset Crash Keep up your momentum on these problems 8P1-3, 7-14 7) 9000 N Read 8.3 8) 2690 N 8CQ9-11, 13 9) −2.40 × 103 𝑁, same 10) 150 kg m/s, 1.43 m/s, 15 m/s Answers 1) 1.50 × 104 102 𝑘𝑔 ⋅ 𝑚/𝑠 𝑚 3) 8 × 104 𝑠 , 1.20 × 106 𝑘𝑔 ⋅ 𝑚 𝑚 , −0.0100 𝑠 𝑠 𝑘𝑔 ⋅ 𝑚/𝑠, 625 times, 6.66 × 2) 1.20 × 108 𝑘𝑔, 1.21 × 103 𝑡𝑖𝑚𝑒𝑠 11) -800 kg m/s, -1.20 m/s 12) 6.67 × 103 𝑁 13) −1.50 × 106 𝑁, −1.00 × 105 𝑁 14) -12.0 m/s, -360 N, 0.300 System All the objects involved in the problem Usually only two objects Internal Forces – Forces that the objects exert on each other External Forces – Forces exerted by things outside of the system Two balls hit in the air During the collision Internal Forces = F12 and F21 External Forces = Weight (W1 and W2) 𝐹Δ𝑡 = 𝑚𝑣𝑓 − 𝑚𝑣0 Object 1: 𝑊1 + 𝐹12 Δt = m1 vf1 − m1 v01 Object 2: 𝑊2 + 𝐹21 Δt = m2 vf2 − m2 v02 Add 𝑊1 + 𝑊2 + 𝐹12 + 𝐹21 Δt = m1 vf1 + m2 vf2 − m1 v01 + m2 v02 𝐸𝑥𝑡 𝐹 + 𝐼𝑛𝑡 𝐹 Δ𝑡 = 𝑝𝑓 − 𝑝0 Since F12 and F21 are equal and opposite Sum of internal forces = 0 External Forces Δ𝑡 = 𝑝𝑓 − 𝑝0 If Isolated system: 0 = 𝑝𝑓 − 𝑝0 OR 𝑝0 = 𝑝𝑓 Law of Conservation of Momentum In an isolated system the total momentum remains constant System can contain any number of objects Watch Crash Video Two billiard balls are colliding on a table. In order to apply the law of conservation of momentum, what should the system be? One ball or both billiard balls? Two billiard balls. External Forces: Weight and Normal Force If the table is horizontal these cancel If it were one ball, then the force of the second ball hitting it would not cancel with anything. A hockey puck of mass 0.17 kg and velocity 5 m/s is caught by a .5 kg mitten laying on the ice. What is the combined velocity after the puck is in the mitten? (ignore friction) v = 1.27 m/s A 5 kg baseball pitching machine is placed on some frictionless ice. It shoots a 0.15 kg baseball horizontally at 35 m/s. How fast is the pitching machine moving after it shoots the ball? -1.05 m/s This is why you feel recoil when you shoot a gun Reasoning Strategy 1. Decide on the system 2. Identify internal and external forces 3. Is the system isolated? If no, then can’t use conservation of momentum 4. Set the total final momentum of the isolated system equal to the total initial momentum Solving problems is fun! 8P23-27 Read 8.4, 8.5 8CQ15-17 Answers 23) 0.122 m/s 24) 0.272 m/s 25) 2690 N, the same 26) 27.4 m/s 27) 22.4 m/s Watch Bumper Video Watch Truck Crash video Subatomic – kinetic energy often conserved Macroscopic – kinetic energy usually not conserved Converted into heat Converted into distortion or damage Elastic – kinetic energy conserved Inelastic – kinetic energy not conserved Completely inelastic – the objects stick together You are playing marbles. Your .10 kg shooter traveling at 1 m/s hits a stationary .05 kg cat’s eye marble. If it is an elastic collision what are the velocities after the collision? vc = 1.33 m/s vs = .333 m/s A ballistic pendulum can be used to determine the muzzle velocity of a gun. A .01 kg bullet is fired into a 3 kg block of wood. The block is attached with a thin .5m wire and swings to an angle of 40°. How fast was the bullet traveling when it left the gun? v = 455 m/s Watch Child Seat video Watch Reducing Risk video Your paper has two dimensions to write on…use them. 8P29-32, 34, 36, 38, 42 Read 8.6, 8.7 8CQ20-22 30) -34.9 m/s, 0.150 m/s 31) -86.4 N, -0.389 J, 64.0% 32) 1.78 m/s, -267 J 34) −1.05 × 10−2 m/s, 1.818 × 108 J 36) 0.182 m/s, 8.52 × 103 J Answers 38) 4.58 m/s, 31.5 J, -0.491 m/s, 3.38 J, Bigger mass = smaller KE 29) 0.25 m/s 42) 24.8 m/s Collisions in 2 dimensions Done the same as in one dimension Do one equation for x and one for y A 2-kg cue ball is moving at 5.00 m/s and collides with a stationary 8 ball of equal mass. After the collision the cue ball moves at 30° to the left of the original direction and the 8-ball moves 90° to the right of the cue balls final direction. Find the velocities after the collision. vc = 4.33 m/s v8 = 2.5 m/s Rocket propulsion Uses Newton’s 3rd Law of Motion Forcing something out back to get an equal force forward As hot gasses are forced out the back, the velocity of the rocket increases. Before: 𝑝 = 𝑚𝑣 After: 𝑚 − Δ𝑚 𝑣 + Δ𝑣 + Δ𝑚 𝑣 − 𝑣𝑒 Impulse: Δ𝑝 = 𝐹Δ𝑡 = −𝑚𝑔Δ𝑡 Impulse = change in momentum −𝑚𝑔Δ𝑡 = 𝑚 − Δ𝑚 𝑣 + Δ𝑣 + Δ𝑚 𝑣 − 𝑣𝑒 − 𝑚𝑣 −𝑚𝑔Δ𝑡 = 𝑚𝑣 + 𝑚Δ𝑣 − Δ𝑚𝑣 − Δ𝑚Δ𝑣 + Δ𝑚𝑣 − Δ𝑚𝑣𝑒 − 𝑚𝑣 −𝑚𝑔Δ𝑡 = 𝑚Δ𝑣 − Δ𝑚𝑣𝑒 Δ𝑣 Δ𝑚𝑣𝑒 −𝑔 = − Δ𝑡 𝑚Δ𝑡 𝑣𝑒 Δ𝑚 −𝑔 =𝑎 𝑚Δ𝑡 Practical limit of 𝑣𝑒 = 2.5 × 103 m/s Δ𝑚 𝑣 Δ𝑡 𝑒 = thrust Final velocity of one stage rocket 𝑣 = 𝑣𝑒 ln 𝑚0 𝑚𝑟 Even though rocket accelerates, the center of mass of the rocket + gases is actually in free fall 87 In order to leave earth 88 of the rocket must be fuel if no air resistance With air resistance 179 must be 180 fuel Try to fix using multistage rockets where the mass of the rocket decreases as the stages are used up What is the acceleration of a 5000-kg rocket taking off from Mars where the acceleration due to gravity is only 3.71 m/s2, if the rocket expels 10 kg of gas per second at an exhaust velocity of 2.25 × 103 m/s? Watch curling video Launch into the work and rocket through these problems. 8P45, 47, 48, 50, 53-56 Answers 45) 3.00 m/s at 60° down 47) -2.26 m/s, 7.63 × 103 J 48) 9.10 m/s at -14.7°, 0.689 50) 8.46 m/s at 33.6°, −8.78 × 104 J 53) 39.2 m/s2 54) 1.92 m/s2 55) 4.16 × 103 m/s 56) 1.60 × 104 𝑚/𝑠, 1.80 × 10−3 m/s2