PHYSICS 220 Exam 1 Oct. 5 at 8:00 PM, a one hour exam Potential Energy and Energy Conservation No equations will be provided but you bring one 8 1/2 X 11 inch crib sheet with whatever you want on it Bring a calculator, #2 pencil, and Purdue ID card 1 Locations by Lab Instructor WTHR 200 - Mandal, Nirajan Lin, Tzu-Ging Jung, Yookyung CL50 224 - Cervantes, Mayra Mukundan, Vineetha Li, Xin-A See Home Page For lab hours Work Done by Gravity Depends only on initial and final height! W g = -mg(y f - y i ) = -mg!y Independent of path If you end up where you began, W g = 0 Define: Potential Energy PE grav = mgy W g = -!PE grav PHYS 114 -- Setty, Chandan Shao, Siyi Zhang, Shunyuan We call this a Conservative Force because we can define a Potential Energy to go with it. Lecture 9 4
Potential Energy Work by Variable Force W = F x!x Work is the area under the F vs x plot Spring: F spring = -k x Work done by gravity is independent of path W g = -mg (y f - y i ) = -!PE g Force F=kx Define PE g = mgy Only the difference in potential energy is physically meaningful, i.e., you have the freedom to choose the reference (or zero potential energy) point. Works for any CONSERVATIVE force 5 Distance Potential Energy: -W=!PE s = 1/2 k x 2 6 Work-Energy with Conservative Forces Work-Energy Theorem " W =!KE Move work by conservative forces to other side " W nc =!KE +!PE If there are NO non-conservative forces 0 =!KE +!PE =!E mech 0 = KE f! KE i + PE f! PE i KE i + PE i = KE f + PE f!w = W cons + W nc = "KE W nc =!KE " W cons E i = E f Conservation of mechanical energy 7 Demo 1M - 03 Three paths all drop same distance Change in PE same independent of path Same gain in KE Same v at end of chute means same trajectory
Skiing Example (no Friction) Skiing with Friction A skier goes down a 78 meter high hill with a variety of slopes. What is the maximum speed the skier can obtain starting from rest at the top? A 50 kg skier goes down a 78 meter high hill with a variety of slopes. She is observed to be going 30 m/s at the bottom of the hill. How much work was done by friction? Conservation of energy: Work Energy Theorem: KEi + PEi = KEf + PEf!m vi 2 Wnc = (KEf + PEf) - (KEi + PEi) + m g yi =! m vf + m g yf 2 = (! m vf2 + m g yf) - (! m vi2 + m g yi) 0 + g y i =! vf 2 + g y f =! (vf2 - g yi )m vf2 = 2 g (yi-yf) = (! (30)2 9.8 x 78) 50 vf = sqrt( 2 g (yi-yf)) = (450 764) 50 Joules = -15700 Joules vf = sqrt( 2 x 9.8 x 78) = 39 m/s 10 Pendulum Exercise B) Zero With no regard for his own personal safety your physics professor will risk being smashed by a bowling ball pendulum! If released from a height h, how far will the bowling ball reach when it returns? C) Negative W = F d cos ". But " = 90 degrees so Work is zero. How fast is the ball moving at the bottom of the path? Conservation of Energy (Wnc=0) Conservation of Energy (Wnc=0) #Wnc =!KE +!PE 0 = KEfinal - KEinitial + PEfinal - PEinitial KEinitial + PEinitial = KEfinal + PEfinal 0 + mgh =! m v2final + 0 vfinal = sqrt(2 g h) #Wnc =!KE +!PE 11 Pendulum Exercise As the pendulum falls, the work done by the string is A) Positive 0 = KEfinal - KEinitial + PEfinal- PEinitial KEinitial + PEinitial = KEfinal + PEfinal 0 + mghinitial = 0 + mghfinal hinitial = hfinal h 12 h 12
Example How high will the pendulum swing on the other side now? A) h 1 > h 2 B) h 1 = h 2 C) h 1 < h 2 Demo 1M - 10 Ball will not fall if N > 0 m h 1 h 2 Conservation of Energy (W nc =0) #W nc =!KE +!PE KE initial + PE initial = KE final + PE final 0 + mgh 1 = 0 + mgh 2 h 1 = h 2 Demo 1M - 02 mv 2 /r = mg + N v 2 > gr mgh i = mgh f +1/2 mv 2 mgh i > mg2r +1/2 mgr h i > 5/2 r 13 Gravitational Potential Energy If the gravitational force is not constant or nearly constant, we have to start from Newton s gravitational force law F = G m 1 m 2 r 2 The gravitational potential energy is: PE g =!G m 1 m 2 r if PE g = 0 for r = " Exercise A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x 1 from its relaxed position while momentarily coming to rest. If the initial speed of the box were doubled and its mass were halved, how far x 2 would the spring compress? A) B) C) x 15 16
Exercise Use the fact that E i = E f In this case, E f = 0 + 1 / 2 kx 2 and E i = 1 / 2 mv 2 + 0 so kx 2 = mv 2 So if v 2 = 2v 1 and m 2 = m 1 /2 m 1 v 1 In the case of x 1 17 x 1 m 1 P av = W /!t Power (Rate of Work) Units: Joules/Second = Watt W = F!r cos" = F (v!t) cos" P = F v cos" How much power does it take for a (70 kg) student to run up the stairs (5 meters) in 7 seconds? P av = W / t = m g h / t = (70 kg) (9.8 m/s 2 ) (5 m) / 7 s = 490 J/s or 490 Watts 18