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Physics 101:  Lecture 10 Potential  Energy & Energy Conservation  ,[object Object],Hour Exam 1:  Today! -(no) cheating! -conflict in  151 , 5:15pm -regular exam – room assignments
Review ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Preview ,[object Object],
Preflight 4 ,[object Object],“ everything”  “nothing” “ conservative vs. nonconservative / Wnc” “ The equations were kind of confusing…” “ Understanding what the symbols mean”  sign of kinetic and potential energies. . “ physics is difficult; it takes too much WORK and ENERGY to understand it.  ” “ don't really understand work...  ” Studying for the exam
Work Done by Gravity 1 ,[object Object],W g   = (mg)(S)cos  S = h W g  = mghcos(0 0 ) = mgh  y = y f -y i  = -h W g  = -mg  y   mg S y x mg S y x Y i  = h Y f  = 0 Y i  = h Y f  = 0
Work Done by Gravity 2 ,[object Object],W g   = (mg)(S)cos  S = h   W g  = mghcos(180 0 ) = -mgh  y = y f -y i  = +h W g  = -mg  y mg S y x Y f  = h Y i  = 0
Work Done by Gravity 3 ,[object Object],W g   = (mg)(S)cos    S = h/cos  W g  = mg(h/cos  )cos  W g  = mgh   y = y f -y i  = -h W g  = -mg  y   h  mg S
Work and Potential Energy ,[object Object],[object Object],[object Object],[object Object],[object Object]
Conservation ACT ,[object Object],[object Object],[object Object],[object Object],[object Object],correct
Skiing Example (no friction) A skier goes down a 78 meter high hill with a variety of slopes. What is the maximum speed she can obtain if she starts from rest at the top? Conservation of energy:  W nc  =   K +   U K i  + U i  = K f  + U f   ½ m v i 2  + m g y i  = ½ m v f 2  + m g y f 0 + g y i  = ½ v f 2  + g y f   v f 2  = 2 g (y i -y f ) v f  = sqrt( 2 g (y i -y f )) v f  =   sqrt( 2 x 9.8 x 78) = 39 m/s 26 0 = K f -K i  + U f  - U i
Pendulum ACT ,[object Object],[object Object],[object Object],W = F d cos   .  But    = 90 degrees so Work is zero. h Conservation of Energy (W nc =0)  W nc  =   K +    U 0  = K final  - K initial  + U final - U initial   K initial  + U initial  = K final +U final 0 + mgh = ½ m v 2 final  + 0 v final  = sqrt(2 g h)
Pendulum Demo ,[object Object],h Conservation of Energy (W nc =0)  W nc  =   K +    U 0  = K final  - K initial  + U final - U initial  K initial  + U initial  = K final +U final 0 + mgh initial  = 0 + mgh final   h initial  = h final
Lecture 10,  Preflight 1  ,[object Object],[object Object],“ they will all have the same speed at the bottom.”   “ The ball being dropped will reach the ground fastest since it doesn't have to travel as far.” 1 2 3 correct
Lecture 10,  Preflight 2 ,[object Object],[object Object],Conservation of Energy (W nc =0)  W nc  =   K +    U K initial  + U initial  = K final +U final 0 + mgh = ½ m v 2 final  + 0 v final  = sqrt(2 g h)  1 2 3 correct
Skiing w/ Friction A 50 kg skier goes down a 78 meter high hill with a variety of slopes. She finally stops at the bottom of the hill. If friction is the force responsible for her stopping, how much work does it do? Work Energy Theorem: W nc  = K f -K i  + U f  - U i   = ½ m v f 2  - ½ m v i 2  + m g y f  – m g y i =  0+0+0 - g y i  m =  – 764 x 50  Joules = -38200  Joules Similar to bob sled homework
Galileo’s Pendulum ACT ,[object Object],[object Object],h 1 h 2 m Conservation of Energy (W nc =0)  W nc  =   K +    U K initial  + U initial  = K final +U final 0 + mgh 1  = 0 + mgh 2   h 1  = h 2
Power (Rate of Work) ,[object Object],[object Object],[object Object],P = W / t =  m g h  / t = (70 kg) (9.8 m/s 2 ) (5 m) / 7 s = 490 J/s  or 490 Watts
Summary ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

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Lecture10

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  • 9. Skiing Example (no friction) A skier goes down a 78 meter high hill with a variety of slopes. What is the maximum speed she can obtain if she starts from rest at the top? Conservation of energy:  W nc =  K +  U K i + U i = K f + U f ½ m v i 2 + m g y i = ½ m v f 2 + m g y f 0 + g y i = ½ v f 2 + g y f v f 2 = 2 g (y i -y f ) v f = sqrt( 2 g (y i -y f )) v f = sqrt( 2 x 9.8 x 78) = 39 m/s 26 0 = K f -K i + U f - U i
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  • 13.
  • 14. Skiing w/ Friction A 50 kg skier goes down a 78 meter high hill with a variety of slopes. She finally stops at the bottom of the hill. If friction is the force responsible for her stopping, how much work does it do? Work Energy Theorem: W nc = K f -K i + U f - U i = ½ m v f 2 - ½ m v i 2 + m g y f – m g y i = 0+0+0 - g y i m = – 764 x 50 Joules = -38200 Joules Similar to bob sled homework
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Editor's Notes

  1. Careful that independent of path really but does depend on initial and final points.