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PHYS_2014_Lecture_12 - Lecture 12 Energy and Work Energy...

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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 12, Slide 1 Energy can be transformed from one form into the other and back again. θ M M h M 0 U mgh K = = 2 0 1 2 U K mv = = 0 U mgh K = = A Pendulum transforms Gravitational Potential Energy to Kinetic Energy and back to Potential Energy Lecture 12: Energy and Work 2 2 1 1 2 2 i i f f mv mgy mv mgy + = + i i f f K U K U + = +
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 12, Slide 2 Energy depends on your Frame of Reference i.e. Energy is always measured relative to some chosen point. 20 m 10 m 0 m m =10kg 1960J g U mgh = = 980J g U mgh = = 0J g U mgh = =
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 12, Slide 3 s F c θ =90 ° How much work is done by the ball at the end of a string, moving in a circle around my hand? cos90 cos90 0 0 c W F s F s W = ⋅Δ = Δ = = ringoperator ringoperator arrowrightnosp arrowrightnosp Forces acting at right angles to the direction of motion do no work!
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 12, Slide 4 Conservative Force: a) The net work done in moving a mass between two points depends only on the location of the points relative to one another, not on the particular path followed. b) The net work done in moving a mass through any closed loop (round- trip) path is zero. 0 F ds = arrowrightnosp arrowrightnosp Nonconservative Force: The net work done in moving a mass between two points depends on the particular path, e.g. due to friction, drag, etc. Δ h Δ h °
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 12, Slide 5 Potential energy is always associated with a conservative force. Friction and Drag (air resistance) are examples of nonconservative forces . The energy lost (work done) by nonconservative forces is lost from the system...usually in the form of heat . The energy lost (work done) by conservative forces is still in the system and can be recovered. Conservative and Nonconservative Forces
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 12, Slide 6 A 50 kg ice skater is gliding along the ice, heading due north at 4.0 m/s. The ice has a small coefficient of static friction that prevents the skater from slipping sideways, but μ k =0. Suddenly, a wind from the northeast exerts a force of 4.0 N on the skater. a) Use work and energy to find the skater’s speed after gliding 100 m in this wind. b) What is the minimum value of μ s that allows her to continue moving straight north? w F arrowrightnosp f F arrowrightnosp s Δ arrowrightnosp y x v 1 m
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Fall 2010 Oklahoma State University PHYS2014: Benton Lecture 12, Slide 7 a) The work/energy theorem can be written: 2 2 .
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