Lect07 - Stuff you asked about 5x pudding Talk more about work go over boxes being pulled on a table by a string connected to a box box hanging off

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Unformatted text preview: Stuff you asked about: 5x pudding Talk more about work. go over boxes being pulled on a table by a string connected to a box box hanging off the table How much of the calculus do we need to know? Why did you change from fading in each equation and circling the important important bits to just throwing all the integrations and whatnot onto one eye-bending page? all of them, especially the ball on the curved slope problem I'm incredibly unfamiliar with vector calculus, and I'd like you to clarify how to to combine vectors with integrals properly. Seriously. It's making me rethink how I've been doing physics in high school and kind of scaring me. The physics of massive amounts of bunnies in a very small box that is moving across a pile of frictionless bunnies with a constant acceleration. I don't know if there is an equation for that... Please put this on the lecture slides in class tomorrow. It's my dream to be dream on it. Physics 211 Lecture 7, Slide 1 Physics Physics 211 Lecture 7 Today's Concepts: Work & Kinetic Energy Physics 211 Lecture 7, Slide 2 Physics The Dot Product Physics 211 Lecture 7, Slide 3 Physics Work Kinetic Energy Theorem The work done by force F as it acts on an object that moves The as between positions r1 and r2 is equal to the change in the and is object’s kinetic energy: W = ΔK W= r2 ∫ F ⋅ dl r1 1 2 K = mv 2 Physics 211 Lecture 7, Slide 4 Physics If the force is constant and the directions aren’t changing then this is very simple to evaluate: F Car W= d r2 F ⋅ dl = F ⋅ d ∫ r1 In this case = Fd since cos(0)=1 Physics 211 Lecture 7, Slide 5 Physics Act A lighter car and a heavier truck, each initially at rest, are pushed with the same constant force F. After both vehicles travel a distance d, which of the following statements is true? (Ignore friction) Car Truck F d F d A) The will have the same velocity B) They will have the same kinetic energy C) They will have the same momentum Physics 211 Lecture 7, Slide 6 Physics r2 F dl ∫ F ⋅d l r1 = ΔK Derivation – not so important Derivation not Concept – very important very dl θ A force pushing over some distance will change the kinetic energy will F Physics 211 Lecture 7, Slide 7 Physics Work done by gravity near the Earth’s surface ath p mg Physics 211 Lecture 7, Slide 8 Physics Work done by gravity near the Earth’s surface WTOT = W1 + W 2 + ... + W N dlN = mg ⋅ dl1 + mg ⋅ dl 2 + ... + mg ⋅ dl N dl1 dl2 mg dy1 dl1 dx1 mg Physics 211 Lecture 7, Slide 9 Physics Work done by gravity near the Earth’s surface WTOT = W1 + W 2 + ... + W N dlN = mg ⋅ dl1 + mg ⋅ dl 2 + ... + mg ⋅ dl N = − mgdy1 − mgdy 2 ... − mgdy N Δy dl1 dl2 = − mg Δ y mg W g = − mg Δ y Physics 211 Lecture 7, Slide 10 Physics Work Kinetic Energy Theorem If there are several forces acting then W is the work done by the If is net (total) force: W NET = Δ K r2 W NET = ∫ ( F1 + F2 + ...) ⋅ dl r1 = r2 r2 ∫ F ⋅ dl + ∫ F 1 r1 2 ⋅ dl1 + ... r1 = W1 + W 2 + ... You can just add up the work done by each force Physics 211 Lecture 7, Slide 11 Physics Preflight Three objects having the same mass begin at the same height, and all move down the same vertical distance H. One falls straight down, one slides down a frictionless inclined plane, and one swings on the end of a string. In which case does the object have the biggest net work done on it by all forces during its motion? H Free Fall A) Free Fall Frictionless incline B) Incline String C) String D) All the same Act Three objects having the same mass begin at the same height, and all move down the same vertical distance H. One falls straight down, one slides down a frictionless inclined plane, and one swings on the end of a string. What is the relationship between their velocities when they reach the bottom? H Free Fall A) Vf > Vi > Vp Frictionless incline String B) Vf > Vp > Vi B) C) Vf = Vp = Vi C) Preflight/Act H Free Fall A) Vf > Vi > Vp Frictionless incline B) Vf > Vp > Vi B) String C) Vf = Vp = Vi C) Only gravity will do work: Wg = mgH = 1/2 mv22 - 1/2 mv12 = 1/2 mv22 vf = v i = v p = 2gH W N ET = Δ K Preflight/Act A car drives up a hill with constant speed. Which statement best describes the total work WTOT done on done the car by all forces as it moves up the hill? A) B) C) WTOT = 0 WTOT > 0 WTOT < 0 v A) velocity is not changing, and therefore the work (which equals change in kinetic energy) is zero. B) The net force and the distance are in the same direction. C) The car is moving against gravity so work is negative. Physics 211 Lecture 7, Slide 15 Physics From Monday From A box sits on the horizontal bed of a moving truck. Static friction between the box and the truck keeps the box from sliding around as the truck drives. μS a If the truck moves with constant accelerating to the left as shown, which of the following diagrams best describes the static frictional force acting on the box: A B C Physics 211 Lecture 7, Slide 16 Physics Preflight/Act Preflight/Act A box sits on the horizontal bed of a moving truck which his accelerating to the left as shown. Static friction between the box and the truck keeps the box from sliding. F μS a D The work done on the box by the static frictional force as the truck moves a distance D is: A) Positive B) Negative B) C) Zero C) Physics 211 Lecture 7, Slide 17 Physics Work done by a spring 0 x1 FS x FS = -kx x2 x1 x1 12 12 W = ∫ F ( x) ⋅ dx = − k ∫ xdx = − kx = − kx1 2 2 x1 0 0 Physics 211 Lecture 7, Slide 18 Physics Act 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) x2 = x1 B) x2 = 2 x1 C) x2 = 2 x1 x1 12 WS = − kx1 2 Act Solution v Wspring m x1 = v k x1 12 = − kx1 2 12 ΔK = − mv 2 If v is doubled and m is halved: is is m x2 = 2v = 2 x1 2k Physics 211 Lecture 7, Slide 20 Physics Work done by gravity in general r2 r2 r2 GM e m ⎛1 1⎞ GM e m W = ∫ F (r ) ⋅ dr = − ∫ = = GM e m ⎜ − ⎟ 2 r r r1 r1 r1 ⎝ r2 r1 ⎠ Physics 211 Lecture 7, Slide 21 Physics Act In Case 1 we send an object from the surface of the earth to a height above the earth surface equal to one earth radius. In Case 2 we start the same object a height of one earth radius above the surface of the earth and we send it infinitely far away. In which case is the magnitude of the work done by the Earth’s gravity on the object biggest? A) Case 1 Case 1 B) Case 2 C) They are the same C) Case 2 Act Solution ⎛1 1⎞ GM e m Case 1: W = GM e m ⎜ − ⎟=− 2 RE ⎝ 2 RE RE ⎠ Same! Case 2: RE ⎛1 1⎞ GM e m = GM e m ⎜ − W ⎟ =− 2 RE ⎝ ∞ 2 RE ⎠ Case 1 2RE Case 2 Physics 211 Lecture 7, Slide 23 Physics ...
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This note was uploaded on 09/22/2011 for the course PHYSICS 211 taught by Professor Selig during the Fall '10 term at University of Illinois, Urbana Champaign.

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