Potential Energy and Energy Conservation PHYSICS 220 Lecture 10 Potential Energy and Energy Conservation Lecture 10 Purdue University, Physics 220

Purdue University, Physics 220 Potential Energy Work done by gravity is independent of path Wg = -mg (yf - yi) = - PEg Define PEg = 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 Careful that independent of path really but does depend on initial and final points. Lecture 10 Purdue University, Physics 220

Purdue University, Physics 220 Work by Variable Force W = Fx Dx Work is the area under the F vs x plot Spring: F spring= -k x Potential Energy: -W=PEs = 1/2 k x2 Lecture 10 Purdue University, Physics 220

Work-Energy with Conservative Forces Work-Energy Theorem Move work by conservative forces to other side If there are NO non-conservative forces Conservation of mechanical energy Lecture 10 Purdue University, Physics 220

Purdue University, Physics 220 iClicker Imagine that you are comparing three different ways of having a ball move down through the same height. In which case does the ball reach the bottom with the highest speed? A) Dropping B) Slide on ramp (no friction) C) Swinging down D) All the same A B C correct Conservation of Energy (Wnc=0) KEinitial + PEinitial = KEfinal + PEfinal 0 + mgh = ½ m v2final + 0 vfinal = sqrt(2 g h) Lecture 10 Purdue University, Physics 220

Skiing Example (no 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? Conservation of energy: KEi + PEi = KEf + PEf ½ m vi2 + m g yi = ½ m vf2 + m g yf 0 + g yi = ½ vf2 + g yf vf2 = 2 g (yi-yf) vf = sqrt( 2 g (yi-yf)) vf = sqrt( 2 x 9.8 x 78) = 39 m/s Lecture 10 Purdue University, Physics 220

Purdue University, Physics 220 Skiing with Friction 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? Work Energy Theorem: Wnc = (KEf + PEf) - (KEi + PEi) = (½ m vf2 + m g yf) - (½ m vi2 + m g yi) = ½ (vf2 - g yi )m = (½ (30)2 – 9.8 x 78) 50 = (450 – 764) 50 Joules = -15700 Joules Lecture 10 Purdue University, Physics 220

Purdue University, Physics 220 Power (Rate of Work) Pav = W / Dt Units: Joules/Second = Watt W = F r cosq = F (v t) cosq P = F v cosq How much power does it take for a (70 kg) student to run up the stairs (5 meters) in 7 seconds? Pav = W / t = m g h / t = (70 kg) (9.8 m/s2) (5 m) / 7 s = 490 J/s or 490 Watts Lecture 10 Purdue University, Physics 220

Purdue University, Physics 220 Pendulum Exercise As the pendulum falls, the work done by the string is A) Positive B) Zero C) Negative How fast is the ball moving at the bottom of the path? W = F d cos q. But q = 90 degrees so Work is zero. Conservation of Energy (Wnc=0) SWnc = DKE + DPE 0 = KEfinal - KEinitial + PEfinal - PEinitial KEinitial + PEinitial = KEfinal + PEfinal 0 + mgh = ½ m v2final + 0 vfinal = sqrt(2 g h) h Lecture 10 Purdue University, Physics 220

Purdue University, Physics 220 iClicker How high will the pendulum swing on the other side now? A) h1 > h2 B) h1 = h2 C) h1 < h2 Conservation of Energy (Wnc=0) SWnc = DKE + DPE KEinitial + PEinitial = KEfinal + PEfinal 0 + mgh1 = 0 + mgh2 h1 = h2 m h1 h2 Lecture 10 Purdue University, Physics 220

Gravitational Potential Energy If the gravitational force is not constant or nearly constant, we have to start from Newton’s gravitational force law The gravitational potential energy is: Lecture 10 Purdue University, Physics 220

Purdue University, Physics 220 Problem: How High? A projectile of mass m is launched straight up from the surface of the earth with initial speed v0. What is the maximum distance from the center of the earth RMAX it reaches before falling back down. RMAX m RE v0 M Lecture 10 Purdue University, Physics 220 6

Purdue University, Physics 220 Problem: How High... All forces are conservative: WNC = 0 KE = -PE And we know: RMAX m RE v0 hMAX M Lecture 10 Purdue University, Physics 220 7

Purdue University, Physics 220 Problem: How High... RMAX m RE v0 hMAX M Lecture 10 Purdue University, Physics 220 8

Purdue University, Physics 220 Escape Velocity If we want the projectile to escape to infinity we need to make the denominator in the above equation zero: We call this value of v0 the escape velocity, vesc Lecture 10 Purdue University, Physics 220 9

Purdue University, Physics 220 Exercise A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 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 x2 would the spring compress ? A) B) C) x Lecture 10 Purdue University, Physics 220

Purdue University, Physics 220 Exercise Use the fact that Ei = Ef In this case, Ef = 0 + 1/2 kx2 and Ei = 1/2 mv2 + 0 so kx2 = mv2 In the case of x1 So if v2 = 2v1 and m2 = m1/2 x1 v1 m1 Lecture 10 Purdue University, Physics 220