Your comments I feel like this could work brilliantly in a nice sized stadium seating lecture hall. Point to a student in the back/upper seats, say they look like they have a lot of potential and ask them to come down front for a demonstration. When they get there say, "No, I've misjudged. You don't have much potential at all." Send them back. When they're half way back up, stop them and say "Well, wait, now I think I see some potential in you again!". Repeat until they catch on or give up. This prelecture was definitely the hardest so far. not awful, but kind of confusing It was fairly simple stuff - it's just that there are SO MANY EQUATIONS. I feel i invest too much time in studying this lecture videos, however dont comprehend more then 60 % of the material? There was so much discussed in this lecture, I am very overwhelmed. I would like all of it to be gone over in class. What speed does it take for something hurtling towards the earth to become a fireball? If mass is irrelevant for a roller coaster negotiating a loop, why are there different height restrictions for rides? cool story bro I think the class should be given potential energy in the form of donuts because they have the potential to be delicious. The relationships between potential and kinetic energy. Also, more AC/DC or the likes. That woke me up instantly.

Gradebook & Midterm & Office Hours Thanks for reminding me to tell you all about the exam: “Prelectures lack examples.. all this conceptual crap is NOT on the exam so... correlate accordingly so my mom doesn't ask why I failed all the midterms and finals because the lectures don't help and the prelectures are just as helpless -__-...” Around half the problems purely conceptual: Understanding concepts is in fact the key to solving all of the problems. Have a look at the practice exams. The first exam is in 2 weeks (Wednesday Feb 22 at 7pm – see planner) If you have a conflict you can sign up for the conflict exam in your grade-book. Don’t forget – we have extra office hours during exam week.

Today's Concepts: a) Potential Energy b) Mechanical Energy Physics 211 Lecture 8 Today's Concepts: a) Potential Energy b) Mechanical Energy

Work done by a Spring Use formula to get the magnitude Use picture to get the sign If you’re worried about the sign: “Can you spend more time explaining the spring problems.” On the last page we calculated the work done by a constant force when the orientation of the path relative to the force was changing. On this page we will calculate the work done by a spring in one dimension. In this case the orientation of the path relative to the force is simple, but the magnitude of the force changes as we move. If the coordinate system is chosen such that x=0 is the relaxed length of the spring as shown, then the force exerted by the spring on some object attached to its end as function of position is given by Hookes law [F = -kx]. As we move the object between two positions x1 and x2, the force on the object clearly changes. Breaking the movement into tiny steps we see that the work done by the spring along each step will depend on the position : [dW = F(x)*dx = -kxdx]. To find the total work done we need to integrate this expression between x1 and x2 [show]. The mathematics of doing this is straightforward, but its very important to realize what it means conceptually. Evaluating a one dimensional integral of some function F(x) between two points x1 and x2 is really just finding the area under the curve between these points [show] . Evaluating the above integral to find the work done is therefore just finding the area under the force versus position plot between the points x1 and x2. [show] Notice that the formula for the work done by a spring that we just derived depends only on the endpoints of the motion x1 and x2, hence we see that this is a conservative force also. 4

Clicker Question A box attached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new location. D During this motion, the spring does: A) Positive Work B) Negative Work C) Zero work

Clicker Question A box attached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new location. D During this motion, your hand does: A) Positive Work B) Negative Work C) Zero work

Clicker Question A box attached at rest to a spring at its equilibrium length. You now push the box with your hand so that the spring is compressed a distance D, and you hold the box at rest in this new location. D During this motion, the total work done on the box is: A) Positive B) Negative C) Zero

Homework Problem Wspring = ½ k x2 = ½ (4111) (0.456)2 = 427 J

Homework Problem Wtot = DK Wspring = 427J = ½ mv2 v2 = 2 * 427 / 12 v = 8.44 m/s

Homework Problem Wfriction = -mmgd Wfriction = - (0.45)(12)(9.81)(2.1) Wfriction = - 111.2 J

Homework Problem Wtot = DK Wspring + Wfriction = 427 J - 111.2 J = ½ mv2 v2 = 2 * (427-111.2) / 12 v = 7.25 m/s

Summary Lecture 7 Work – Kinetic Energy theorem Lecture 8 For springs & gravity (conservative forces) Total Mechanical Energy E = Kinetic + Potential Work done by any force other than gravity and springs will change E

Relax. There is nothing new here It’s just re-writing the work-KE theorem: In P211 everything except gravity and springs is NC Define Mechanical Energy: If other forces aren't doing work “So what exactly is the point of potential energy then? Also, why is 'U' the character used to represent it?”

Homework Problem DE = WNC DUspring + ½ mv2 = Wfriction -427 J + ½ mv2 = - 111.2 v2 = 2 * (427-111.2) / 12 v = 7.25 m/s

Finding the potential energy change: Use formulas to find the magnitude Check the sign by understanding the problem…

h CheckPoint A) v1 > v2 > v3 B) v3 > v2 > v1 Three balls of equal mass are fired simultaneously with equal speeds from the same height h above the ground. Ball 1 is fired straight up, ball 2 is fired straight down, and ball 3 is fired horizontally. Rank in order from largest to smallest their speeds v1, v2, and v3 just before each ball hits the ground. 1 2 3 h A) v1 > v2 > v3 B) v3 > v2 > v1 C) v2 > v3 > v1 D) v1 = v2 = v3 “They have the same U and K to start with and delta_U is the same” “I'm so confused- didn't we cover this in when we went over kinematics?”

CheckPoint 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, how far x2 would the spring compress? A) B) C) x

CheckPoint x A) B) C) A) Because w is proportional to x^2, I thought the doubled speed would result in square root of x. B) Both velocity and the distance compressed are squared so they will changes with a 1 to 1 ratio C) because the v is squared, therefore doubling it, quadruples the result

Vertical Springs They both look the same !! kx2 M x DU = ½ kd2 - mgd = ½ ky’2 Equilibrium without mass d Equilibrium with mass They both look the same !!

From Tuesday’s lecture “Why did you lie about the apple?” I didn’t - we can explain this now: DE = WNC DK + DU Whand DU = Whand The change in potential energy of the apple is due to just the work done on it by your hand as you lift it.

From Tuesday’s lecture “Why did you lie about the apple?” The Work – Kinetic Energy theorem says the same thing: Wtot = DK Whand + Wgravity = 0 Whand = - Wgravity Whand = DUgravity Same result: The change in potential energy of the apple is due to just the work done on it by your hand as you lift it.

CheckPoint A) K2 = 2K1 B) K2 = 4K1 C) K2 = 4K1/3 D) K2 = 3K1/2 Case 1 In Case 1 we release an object from a height above the surface of the earth equal to 1 earth radius, and we measure its kinetic energy just before it hits the earth to be K1. In Case 2 we release an object from a height above the surface of the earth equal to 2 earth radii, and we measure its kinetic energy just before it hits the earth to be K2. Compare K1 and K2. A) K2 = 2K1 B) K2 = 4K1 C) K2 = 4K1/3 D) K2 = 3K1/2 wrong Case 1 Case 2

Lets look at U… +U0 RE 2RE 3RE For gravity: “Why would you drop a ball from the height 2x the Earth's radius? Seems impractical.”

A) B) C) RE Clicker Question GMem GMem GMem Usurface = Usurface = RE What is the potential energy of an object of on the earths surface: GMem A) Usurface = GMem B) Usurface = RE GMem C) Usurface = 2RE RE

RE Clicker Question A) B) C) Case 1 Case 2 What is the potential energy of a object starting at the height of Case 1? A) B) C) Case 1 Case 2 RE 26

RE Clicker Question A) B) C) Case 1 Case 2 What is the potential energy of a object starting at the height of Case 2? A) B) C) Case 1 Case 2 RE 27

RE A) B) What is the change in potential in Case 1? Case 1 Case 2 28

RE A) B) What is the change in potential in Case 2? Case 1 Case 2 29

A) 2 B) 4 C) 4/3 D) 3/2 What is the ratio Case 1 Case 2

Checkpoint A) K2 = 2K1 B) K2 = 4K1 C) K2 = 4K1/3 D) K2 = 3K1/2 Case 1 In Case 1 we release an object from a height above the surface of the earth equal to 1 earth radius, and we measure its kinetic energy just before it hits the earth to be K1. In Case 2 we release an object from a height above the surface of the earth equal to 2 earth radii, and we measure its kinetic energy just before it hits the earth to be K2. Compare K1 and K2. A) K2 = 2K1 B) K2 = 4K1 C) K2 = 4K1/3 D) K2 = 3K1/2 Case 1 Case 2 31