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Hour Exam #1 Kinetic energy • The kinetic energy of an object is • • • • • Hour Exam I, Wed. Feb. 14, in-class (50 minutes) Material Covered: Chap 1, 3-6 One page of notes (8.5” x 11”) allowed 20 multiple choice questions Scantron sheets will be used bring #2 HB pencils and calculator • In-class review, Monday Feb. 12 1 2 mv 2 • Positive work done on isolated object – object speeds up, kinetic energy increases • Negative work ! done on isolated object – object slows down, kinetic energy decreases On-line review questions, previous exams, at course web site uw.physics.wisc.edu/~rzchowski/phy107 Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 1 Fri, Feb. 9 Work-Energy relation 1 gallon gasoline has energy content of 125,000 BTU 1 BTU ~ 1,000 J (actually 1,055.056 J) Suppose the energy in gasoline was converted completely into kinetic energy of a 1000 kg car. How fast would the car be going after burning 1 gallon of gas? A. 5 m/s B. 10 m/s 1 125,000,000 J = (1000 kg)v 2 C. 50 m/s 2 D. 100 m/s v 2 = 250,000 m 2 /s2 E. 500 m/s v = 500 m /s = 1,118 miles /hour! • As the result of a net work W net, the velocity increases to vfinal, 1 2 ! and the Kinetic Energy increases to 2 mv final 1 1 2 W net = mv 2final " mv initial 2 !2 The change in kinetic energy of an isolated object is equal to the net work done on it. ! Phy107, Spring 2007 Lecture 9 2 Energy content • Suppose object initially moving at vinitial 2 It has kinetic energy 1 mv initial 2 Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 3 Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 4 ! Friction and resistance Storing energy 1N Move ball up at constant speed -> I exert force exactly canceling gravity • Means some of the energy went elsewhere – Air resistance – Friction of tires on road – Heat losses in engine I exert force mg on ball. Over a distance h, I have done mgh of work • Total energy is always conserved 1N Energy is stored – Some of it just left gas-car system h Lifting force mg Can be released at any time Called ‘potential energy’ Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 5 Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 Gravitational force -mg 6 1 Potential energy Releasing potential energy • Let go of the ball • The potential energy of a system is the work required to get the system into that configuration. – i.e. turn off the upward force I apply • Gravity force accelerates ball downward • Examples – For a pendulum, it is the work required to move the bob to the top of its swing. – For a falling apple, it is the work required to lift the apple. – For a spring, it is the work required to compress the spring • Gravity does work on the ball – Force = mg directed downward – Work done by gravity = Force x Distance = mg x h = mgh 1 mgh = mv 2final 2 Work turned into kinetic energy Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 7 Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 8 ! Gravity and work Question A ball is dropped from 5 meters. How fast is it going when it hits the ground? • Work done to move object vertically up work done is the same for both paths • The work done is ‘independent of path’ • Change in gravitational energy, A. 1 m/s B. 2 m/s Change in energy = mgh C. 5 m/s D. 10 m/s true for any path : h is the height difference, yfinal - yinitial • A falling object converts gravitational potential energy to kinetic energy 2 1 mgh = mv 2final v final = 2gh " v final = 2gh 2 = 2 #10m /s2 # 5m = 100m 2 /s2 = 10m /s ! Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 9 Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 10 ! Question Potential E independent of path • Gravitational force pointed directly downward – only the vertical distance determines the potential energy. If balls are released from the same height on these two different tracks, which will have greater speed at the bottom? A. Both same B. Track A C. Track B • We say it is ‘independent of the path’ (also called ‘conservative’) • True for most ‘non-contact’ (field) forces. – Gravity – Electromagnetism – Nuclear forces Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 B A 11 Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 12 2 Energy conservation Testing conservation of energy • At top: • Total energy ‘E’ is sum of – Kinetic energy =0 1) K = kinetic energy (visible) 2) U = potential (invisible) energy – Potential energy = mgh • At bottom: – Kinetic energy = 1/2 m v2 – Potential energy = 0 • E = K + U = constant • On flat section, use timer and distance traveled to determine speed. • Many situations become much clearer from an energy perspective. Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 13 Fri, Feb. 9 h Phy107, Spring 2007 Lecture 9 A pendulum 14 Work At top of swing, velocity of ball is zero, so it’s kinetic energy is zero. I do mgh of work on the bowling ball. Gravity did -mgh of work on the ball. Net work = 0. No change in kinetic energy. At the bottom of the swing, it’s velocity is very large, so it’s kinetic energy is large. How much work was done on the Earth? None - the Earth did not move. Work = Force x Distance Final position Where did this energy come from? θ θ h h Initial position Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 15 Now release the ball Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 16 Questions about the pendulum • Energy stored in the system as potential energy. • Releasing the ball lets it accelerate and turn the potential energy into kinetic energy. top of swing h h bottom of swing Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 17 Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 18 3 Question Question If the pendulum swing has 1.0 m vertical height, what is the kinetic energy of the 1 kg pendulum bob at the top of its swing? A. 1 J B. 10 J C. 0.1 J D. 1.4 J The velocity is zero, so the E. 0 J If the pendulum swing has 1.0 m vertical height, what is the kinetic energy of the 1 kg pendulum bob at the bottom of its swing? A. 1 J B. 0.1 J C. 1.4 J D. 10 J Conservation of energy. Kinetic = E. 0 J kinetic energy is zero. Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 potential, potential = mgh = (1kg)x(10m/s2)x(1m)=10 J 19 Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 Conservation of energy 20 What speed You are jumping a 600 kg motorcycle between ramps. The ramps are 10 meters high. In order to jump successfully, the bike must leave the ramp with a speed of 10 m/s. (Kinetic En.=30,000 J) What kinetic energy must the motorcycle have entering the jump? What speed must the 600 kg motorcycle have at the ramp entrance (needs 90,000 J)? A. B. C. D. A. 30,000 J B. 60,000 J C. 90,000 J 10 m/s 17.3 m/s 13.2 m/s 5.0 m/s = 39 mi/hr (1/2)mv2 = 90,000 J (300 kg)x(v)2 = 90,000 J v = sqrt(300) m/s = 17.32 m/s Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 21 Fri, Feb. 9 Power P= Work Joules(J) , " Watts (W) time second(s) You run up the first flight of stairs, then walk up the second flight. How do the work and power compare on the two flights? A. B. C. D. It is measured in Watts. (also Horsepower, 1 horsepower = 750 Watts) Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 22 Question Power is the rate at which work is done ! Phy107, Spring 2007 Lecture 9 23 Work Work Work Work Fri, Feb. 9 same, power same different, power same different, ` power different same, power different Phy107, Spring 2007 Lecture 9 24 4 Question Question You press a 500 N weight up to arms length (0.8m) in 2 seconds. Your power output while lifting was You weigh 500 N and run up a stairs with a 4 m vertical height. You do this in 5 seconds. Your power output is A. B. C. D. A. B. C. D. 400 400 100 200 Fri, Feb. 9 J W W W Power = work / time = Force x Distance / time = (500N)x(0.8m)/2 sec = 400 J / 2 sec = 200 Watts Phy107, Spring 2007 Lecture 9 25 400 W 200 W 100 W 50 W Weight = 500 N = mg Energy change = (mg)h = 2,000 J Power = energy / time = 400 W Fri, Feb. 9 Phy107, Spring 2007 Lecture 9 26 5