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sy13_oct17_07hc - Physics 207 Lecture 13 Oct 15 Physics...

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Physics 207: Lecture 12, Pg 1 Physics 207, Physics 207, Lecture 13, Oct. 15 Lecture 13, Oct. 15 Agenda: Finish Chapter 10, start Chapter 11 Agenda: Finish Chapter 10, start Chapter 11 Assignment: Assignment: HW5 due tonight HW5 due tonight HW6 available today HW6 available today Monday, finish reading chapter 11 Monday, finish reading chapter 11 Chapter 10: Energy Chapter 10: Energy Potential Energy (gravity, springs) Kinetic energy Mechanical Energy Conservation of Energy Start Chapter 11, Work
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Physics 207: Lecture 12, Pg 2 Chapter 10: Energy Chapter 10: Energy Rearranging Newton’s Laws gives (Fd vs . ½ mv 2 relationship) -2mg (y f y i ) = m (v yf 2 - v yi 2 ) or ½ m v yi 2 + mgy i = ½ m v yf 2 + mgy f and adding ½ m v xi 2 + ½ m v zi 2 and ½ m v xf 2 + ½ m v zf 2 ½ m v i 2 + mgy i = ½ m v f 2 + mgy f where v i 2 = v xi 2 +v yi 2 + v zi 2 ½ m v 2 terms are referred to as kinetic energy
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Physics 207: Lecture 12, Pg 3 Energy Energy If only “conservative” forces are present, the total energy If only “conservative” forces are present, the total energy ( ( sum of potential, U, and kinetic energies, K sum of potential, U, and kinetic energies, K ) of a system ) of a system is is conserved conserved . K ≡ ½ mv 2 U ≡ mgy K and U may change, but E = K + U mech remains constant. E mech = K + U = constant constant E mech is called “mechanical energy” K i + U i = K f + U f
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Physics 207: Lecture 12, Pg 4 Another example of a conservative system: Another example of a conservative system: The simple pendulum. The simple pendulum. Suppose we release a mass m from rest a distance h 1 above its lowest possible point. What is the maximum speed of the mass and where does this happen ? To what height h 2 does it rise on the other side ? v h 1 h 2 m
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Physics 207: Lecture 12, Pg 5 Example: The simple pendulum. Example: The simple pendulum. y y=0 y= h 1 What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum.
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Physics 207: Lecture 12, Pg 6 Example: The simple pendulum. Example: The simple pendulum. v h 1 y y= h 1 y=0 What is the maximum speed of the mass and where does this happen ? E = K + U = constant and so K is maximum when U is a minimum E = mgh 1 at top E = mgh 1 = ½ mv 2 at bottom of the swing
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Physics 207: Lecture 12, Pg 7 Example: The simple pendulum. Example: The simple pendulum. y y= h 1 =h 2 y=0 To what height h2 does it rise on the other side? E = K + U = constant and so when U is maximum again (when K = 0) it will be at its highest point.
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