Lecture 24 - Lecture 23 New concepts dW (t ) instantaneous...

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Lecture 23
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New concepts displacement of the Force at t instantaneous work done by force ime particle between and ( ) . () t t F t dt dW t F t dr Total Work done by a force in time to , ) ( ) ( ) ( ) ( final initial initial ini t initial f tia inal l t final initial final t t t dW t dW t d Wt W t dt dW t    Kinetic Energy of a par e 2 2 ticl 1 ( 2 . 2 . ) 1 Kt KE m v t mv
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Work-energy theorem , final initial total allforces KW K  Kinetic energy of the particle l at t final Kinetic energy of the particle at t initial Work done by all external forces on the particle between t initial and t final . K E W    Alternative form of writing the work- energy theorem: “The change in the kinetic energy of a particle is equal to the total work done on it by all external forces” 0 . 0 W K E      Positive external work leads to increase in K.E. Since K.E. is always a positive quantitiy, increase in K.E. means increase in speed We say that the particle has “gained” K.E. 0 . 0 W K E      Negative external work leads to decrease in K.E. Since K.E. is always a positive quantitiy, decrease in K.E. means decrease in speed We say that the particle has “lost” K.E.
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Example: Spring force l Equilibrium position 2 1 2 W kl   Y X l Y X 2 1 2 W kl   l Y X ? W 1 2 0 ˆˆ ( )].[ ( )] ( ) ( ) (Limit 0) 1 2 [ total i i i i i i i i l xkx t xdx t k x t dx t t t dx kx kl W       When the work done by the spring force is negative the mass slows down; when it is positive, the mass speeds up
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Example: free fall (gain in K.E.) An object is released from rest and falls through a distance l under gravity. What is its speed v ? • Apply work-energy theorem • What is the initial kinetic energy? K i = 0 • What is the final kinetic energy? K f = (1/2) mv 2 • What is the total work done by all forces during this interval? The only force is gravity, so … W total = mgl • Change in Kinetic energy?
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Lecture 24 - Lecture 23 New concepts dW (t ) instantaneous...

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