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UW-Madison PHYSICS 207 - Definition of Work, The basics

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Lecture 16Definition of Work, The basicsNet Work: 1-D Example (constant force)Net Work: 1-D 2nd Example (constant force)Work: “2-D” Example (constant force)Exercise Work in the presence of friction and non-contact forcesWork and Varying Forces (1D)Slide 9Compare work with changes in potential energyConservative Forces & Potential EnergyA Non-Conservative Force, FrictionA Non-Conservative ForceSlide 14Conservative Forces and Potential EnergyEquilibriumWork & Power:Slide 18Work & Power: ExampleExercise Work & PowerSlide 21Chap. 12: Rotational DynamicsSlide 23Rotational VariablesSystem of Particles (Distributed Mass):For these two particlesSlide 27Slide 28Rotation & Kinetic Energy...Calculating Moment of InertiaCalculating Moment of Inertia...Slide 32Physics 207: Lecture 16, Pg 1Lecture 16•Chapter 11 (Work)Chapter 11 (Work) Employ conservative and non-conservative forces Relate force to potential energy Use the concept of power (i.e., energy per time)Assignment: HW7 due Tuesday, Nov. 1st For Monday: Read Chapter 12, (skip angular momentum and explicit integration for center of mass, rotational inertia, etc.)Exam 2 Exam 2 7:15 PM Thursday, Nov. 37:15 PM Thursday, Nov. 3thth •Chapter 12 Chapter 12  Define rotational inertia Define rotational kinetic energyPhysics 207: Lecture 16, Pg 2Definition of Work, The basicsIngredients: Force ( F ), displacement (  r ) rrdisplacementFFWork, W, of a constant force F acts through a displacement  r :W = F · r (Work is a scalar)(Work is a scalar)Work tells you something about what happened on the path!Did something do work on you? Did you do work on something?Physics 207: Lecture 16, Pg 4Net Work: 1-D Example (constant force)Net Work is F x = 10 x 5 N m = 50 J= 10 x 5 N m = 50 J1 N m ≡ 1 Joule (energy)Work reflects positive energy transferxA force F = 10 N pushes a box across a frictionless floor for a distance x = 5 m.F = 0° StartFinishPhysics 207: Lecture 16, Pg 5Net Work: 1-D 2nd Example (constant force)Net Work is F x = -10 x 5 N m = -50 J= -10 x 5 N m = -50 JWork again reflects negative energy transferxFA force F = 10 N is opposite the motion of a box across a frictionless floor for a distance x = 5 m. = 180° StartFinishPhysics 207: Lecture 16, Pg 6Work: “2-D” Example (constant force)(Net) Work is Fx x = F cos(= F cos(-45°) x = 50 x 0.71 Nm = 35 J = 50 x 0.71 Nm = 35 J xFAn angled force, F = 10 N, pushes a box across a frictionless floor for a distance x = 5 m and y = 0 m  = -45° StartFinishFxPhysics 207: Lecture 16, Pg 7ExerciseWork in the presence of friction and non-contact forcesA. 2B. 3C. 4D. 5A box is pulled up a rough ( > 0) incline by a rope-pulley-weight arrangement as shown below. How many forces (including non-contact ones) are doing work on the box ? Of these which are positive and which are negative? State the system (here, just the box) Use a Free Body Diagram Compare force and path vPhysics 207: Lecture 16, Pg 8Work and Varying Forces (1D)Consider a varying force F(x)FxxxArea = Fx xF is increasingHere W = F · r becomes dW = F dx fixxdxxFW )(Physics 207: Lecture 16, Pg 9•How much will the spring compress (i.e. x = xf - xi) to bring the box to a stop (i.e., v = 0 ) if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ?xvomtoFspring compressedspring at an equilibrium positionV=0tmfixxdxxFW )(boxfixxdxkxW box2021221v mxk Example: Work Kinetic-Energy Theorem with variable forcePhysics 207: Lecture 16, Pg 10Compare work with changes in potential energyConsider the ball moving up to height h(from time 1 to time 2) How does this relate to the potential energy? Work done by the Earth’s gravity on the ball)W = F - x = (-mg) -h = mghU = Uf – Ui = mg 0 - mg h = -mg hU = -WhmgmgPhysics 207: Lecture 16, Pg 11Conservative Forces & Potential EnergyIf a conservative force F we can define a potential energy function U:The work done by a conservative force is equal and opposite to the change in the potential energy function.(independent of path!)W = F ·dr = - U rirf Uf UiPhysics 207: Lecture 16, Pg 12A Non-Conservative Force, FrictionLooking down on an air-hockey table with no air flowing ( > 0). Now compare two paths in which the puck starts out with the same speed (Ki path 1 = Ki path 2) .Path 2Path 1Physics 207: Lecture 16, Pg 13A Non-Conservative ForcePath 2Path 1Since path2 distance >path1 distance the puck will be traveling slower at the end of path 2. Work done by a non-conservative force irreversibly removes energy out of the “system”. Here WNC = Efinal - Einitial < 0  and reflects EthermalPhysics 207: Lecture 16, Pg 14A. U  K B. U  EThC. K  UD. K EThE. There is no transformation because energy is conserved.A child slides down a playground slide at constant speed. The energy transformation isPhysics 207: Lecture 16, Pg 15Conservative Forces and Potential EnergySo we can also describe work and changes in potential energy (for conservative forces)U = - WRecalling (if 1D)W = Fx xCombining these two,U = - Fx xLetting small quantities go to infinitesimals,dU = - Fx dxOr,Fx = -dU / dxPhysics 207: Lecture 16, Pg 16EquilibriumExample Spring: Fx = 0 => dU / dx = 0 for x=xeqThe spring is in equilibrium positionIn general: dU / dx = 0  for ANY function establishes equilibriumstable equilibriumunstable equilibriumUUPhysics 207: Lecture 16, Pg 17Work & Power:Two cars go up a hill, a Corvette and a ordinary Chevy Malibu. Both cars have the same mass. Assuming identical friction, both engines do the same amount of work to get up the hill.Are the cars essentially the same ?NO. The Corvette can get up the hill quickerIt has a more powerful engine.Physics 207: Lecture 16, Pg 18Work & Power:Power is the rate, J/s, at which work is done.Average Power is, Instantaneous Power is,If force constant, W= F x = F (v0 t + ½ at2)and P = W / t = F (v0 + at) tWPdtdWP Physics 207: Lecture 16, Pg 19Work & Power: ExampleAveragePower:A person, mass 80.0 kg, runs up 2 floors (8.0 m) at constant speed. If they climb it in 5.0 sec, what is the average power used?Pavg = F h / t = mgh /  t = 80.0 x 9.80 x 8.0 / 5.0 WP = 1250 W


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UW-Madison PHYSICS 207 - Definition of Work, The basics

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