Lecture15-2A

# Lecture15-2A - Today's Lecture Lecture 15: Chapter 8,...

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Today Today s Lecture s Lecture Lecture 15: Chapter 8, Review of Energy Diagrams Review of Work-Energy Thm Chapter 10, System of Particles

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Energy Diagram Energy Diagram - - Typical Diatomic Molecule Typical Diatomic Molecule If E > 0 , then the atoms cannot approach closer than r = a . The atoms can approach arbitrarily large separations – the molecule is not bound. If E < 0 , then the atoms are trapped between the turning points at b and d . The atoms form a bound system. The equilibrium separation occurs at r = c where dU/dr = 0 . What is the force between the atoms at r = c ?
Energy Diagram Energy Diagram - - Typical Diatomic Molecule Typical Diatomic Molecule Near the bottom of the potential well, r = r c the potential energy is given approximately by: U U 0 a r r c 2 , U 0 7.6 10 19 J , a 2.86 10 16 J / nm 2 , r c .0741 nm . What range of atomic separations is allowed if the total energy is E = -7.17x10 -19 J ? Kinetic energy vanishes at the turning points, E = U there. Solving for r – r c : Hence: r r c  E U 0 / a .44 10 /2.86 10 1.24 10 2 nm r min .0124 .0617 nm r max .0865 nm

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Energy Conservation on a Track Energy Conservation on a Track If a ball is released from a height h on a frictionless track find the expression for the ratio of the time spent on either side during the oscillatory motion. From the conservation of energy the ball will rise to the same height on either side. However the accelerations will be different! The velocities at the bottom are the same, while their respective accelerations (uniform) are a i = g sin q i . Hence: v bot a i t i t 1 t 2 a 2 a 1 g sin 2 g 1 2 1 As expected the side with the steeper angle has the shorter time. Kinematics could also solved this problem, albeit more complicated.
Skier and Two Peaks Skier and Two Peaks If a skier starts from rest at the top of the left-hand peak, what is the maximum value of the coefficient of kinetic friction, m k , that would allow the skier to reach the second peak? From the work-energy theorem: The kinetic energy is zero at both peaks. This means that the loss in potential energy (height) is dissipated into friction. With the use of some trigonometry: Δ K Δ U W nc mg Δ h k cos 1 d 1 k cos 2 d 2 Δ h k d 1 cos 1 d 2 cos 2 k h 1 cot 1 h 2 cot 2

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Skier and Two Peaks Skier and Two Peaks If a skier starts from rest at the top of the left-hand peak, what is the maximum value of the coefficient of kinetic friction, m k , that would allow the skier to reach the second peak? Solving for m k : To reiterate, we could have found the acceleration down the slopes and used kinematics to find this result. However, the work-energy theorem greatly facilitated obtaining the solution.
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## This note was uploaded on 04/16/2008 for the course PHYS PHYS2A taught by Professor Hicks during the Spring '08 term at UCSD.

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Lecture15-2A - Today's Lecture Lecture 15: Chapter 8,...

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