# Lect12 - Physics 211 Lecture 12 Today's Agenda Problems...

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Physics 211: Lecture 12, Pg 1 Physics 211: Lecture 12 Physics 211: Lecture 12 Today's Agenda Today's Agenda z Problems using work/energy theorem ¾ Spring shot ¾ Escape velocity ¾ Loop the loop ¾ Vertical springs z Definition of Power, with example

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Physics 211: Lecture 12, Pg 2 Problem: Spring Shot Problem: Spring Shot z A sling shot is made from a pair of springs each having spring constant k . The initial length of each spring is x 0 . A puck of mass m is placed at the point connecting the two springs and pulled back so that the length of each spring is x 1 . The puck is released. What is its speed v after leaving the springs? (The relaxed length of each spring is x r ). x r x 1 m x 0 m m v
Physics 211: Lecture 12, Pg 3 Problem: Spring Shot Problem: Spring Shot z Only conservative forces are at work, so K+U energy is conserved. E I = E F K = - U s () ( ) 2 r 1 2 r 0 2 r 1 2 r 0 s x x x x k x x x x k 2 1 2 U = = x 0 x 1 m m

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Physics 211: Lecture 12, Pg 4 Problem: Spring Shot Problem: Spring Shot z Only conservative forces are at work, so K+U energy is conserved. E I = E F K = - U s Km v = 1 2 2 m m at rest v
Physics 211: Lecture 12, Pg 5 Problem: Spring Shot Problem: Spring Shot z Only conservative forces are at work, so K+U energy is conserved. E I = E F K = - U s () ( ) 2 r 1 2 r 0 2 x x x x k mv 2 1 = Spring Shot v mm

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Physics 211: Lecture 12, Pg 6 Problem: How High? Problem: How High? z A projectile of mass m is launched straight up from the surface of the earth with initial speed v 0 . What is the maximum distance from the center of the earth R MAX it reaches before falling back down. R MAX R E v 0 m M
Physics 211: Lecture 12, Pg 7 Problem: How High. .. Problem: How High. .. z All forces are conservative: ¾ W NC = 0 ¾ K = - U z And we know: 2 0 mv 2 1 K = = MAX E R 1 R 1 GMm U R MAX v 0 m h MAX R E M = MAX E 2 0 R 1 R 1 GMm mv 2 1

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Physics 211: Lecture 12, Pg 8 Problem: How High. .. Problem: How High. .. MAX E E 2 0 MAX E E MAX E E 2 E MAX E 2 0 MAX E 2 0 R R 1 gR 2 v R R 1 gR 2 R R 1 R R GM 2 R 1 R 1 GM 2 v R 1 R 1 GMm mv 2 1 = = = = = R MAX R E v 0 m h MAX M R R v gR MAX E E = 1 2 0 2
Physics 211: Lecture 12, Pg 9 Escape Velocity Escape Velocity R R v gR MAX E E = 1 2 0 2 z If we want the projectile to escape to infinity we need to make the denominator in the above equation zero: 1 2 0 0 2 −= v gR E v gR E 0 2 2 1 = vg R E 0 2 = We call this value of v 0 the escape velocity, v esc

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Physics 211: Lecture 12, Pg 10 Escape Velocity Escape Velocity z Remembering that we find the escape velocity from a planet of mass M p and radius R p to be: (where G = 6.67 x 10 -11 m 3 kg -1 s -2 ).
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Lect12 - Physics 211 Lecture 12 Today's Agenda Problems...

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