Physics 1A, Section 6

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# Physics 1A, Section 6 - PowerPoint PPT Presentation

Physics 1A, Section 6. October 30, 2008. Section Business. Section 6 for Monday, Nov. 3 is canceled . Visit another section if you wish. Office hour for Nov. 4 is unchanged. Look for graded Quiz 2 in Section 6 box in Bridge, noon Monday at the earliest. Quiz Problem 49. Quiz Problem 49.

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### Physics 1A, Section 6

October 30, 2008

• Section 6 for Monday, Nov. 3 is canceled.
• Visit another section if you wish.
• Office hour for Nov. 4 is unchanged.
• Look for graded Quiz 2 in Section 6 box in Bridge, noon Monday at the earliest.
Quiz Problem 49
• {[k3 + (k1-1 + k2-1)-1]-1 + k4-1]-1
Energy Conservation
• Energy is conserved: K + U + heat + … = constant
• Sometimes, mechanical energy is conserved:
• K + U = mechanical energy = constant
• Example: ½ mv2 + mgh = constant
• This often allows a quick solution of a difficult problem.
Energy Conservation
• Energy is conserved: K + U + heat + … = constant
• Sometimes, mechanical energy is conserved:
• K + U = mechanical energy = constant
• Example: ½ mv2 + mgh = constant
• This often allows a quick solution of a difficult problem.
• However, in other cases, mechanical energy is not conserved, so K + U  constant:
• friction: Energy is lost to heat.
• inelastic collision: Energy is lost to heat.
• This is the same thing as saying the force can’t be described by a potential energy; the force is a function of some variable other than position.
Energy Conservation
• Energy is conserved: K + U + heat + … = constant
• Sometimes, mechanical energy is conserved:
• K + U = mechanical energy = constant
• Example: ½ mv2 + mgh = constant
• This often allows a quick solution of a difficult problem.
• However, in other cases, mechanical energy is not conserved, so K + U  constant:
• friction: Energy is lost to heat.
• inelastic collision: Energy is lost to heat.
• This is the same thing as saying the force can’t be described by a potential energy; the force is a function of some variable other than position.
• In some of those cases, one can resort to using the force to calculate the energy added to the system:

energy input = W = ∫F•ds

Quiz Problem 50
• v = sqrt(2gh)
• F = -kx – mmg, to the right
• W = -kxs2/2 – mmgxs
• Wf = 2mmgxs
• h’ = h – 2mxs
• xs = [-mmg + sqrt(m2m2g2+2kmgh)]/k

Monday, November 3:

Thursday, November 6:
• Section 6 is canceled.
• Tuesday office hour as usual.
• Quiz Problem 38 (momentum/collisions)