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More Constant Acceleration and  Relative Velocity Physics 101:   Lecture 07 Exam I ,[object Object],[object Object],[object Object]
Some Charts
Top  ~Ten Comments… ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Last Time ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Today ,[object Object],[object Object],[object Object]
Pulley, Incline and 2 blocks N A block of mass m 1  = 2.6 kg rests upon a frictionless  incline as shown and is connected to mass m 1  via a flexible cord over an ideal pulley. What is the acceleration of block m 1  if m 2  = 2.0 kg?  X – direction   F x  = m a x : Block 1: T – m 1 g sin(30) = m 1  a 1x   T =   m 1 g sin(30) + m 1  a 1x   Y – direction  F y  = m a y : Block 2: T – m 2  g = m 2  a 2y Note:  a 1x  = - a 2y Combine T   – m 2  g = m 2  a 2y m 1 g sin(30) + m 1  a 1x   – m 2  g = m 2  a 2y m 1 g sin(30) + m 1  a 1x   – m 2  g = -m 2  a 1x m 1  a 1x   +m 2  a 1x = m 2  g - m 1 g sin(30)  (m 1 +m 2 ) a 1x = g (m 2  - m 1  sin(30)) 1.49 m/s 2 T m 2 g m 1 g T y x y x
Newton’s Third Law ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],F finger  box F box  finger
Newton’s 3 rd  Law ,[object Object],[object Object],2) Compare the magnitudes of the acceleration you experience,  a A , to the magnitude of the acceleration of the spacecraft,  a S , while you are   pushing: 1. a A  = a S 2. a A  > a S 3. a A  < a S   a=F/m F same    lower mass give larger a Third Law! correct correct
Newton’s 3 rd  Example ,[object Object],M 1 M 2 T X-direction:  F = ma Block 2:  f 21  = m 2  a 2 Block 1:  T – f 12  = m 1  a 1 N3L says  |f 12 | = |f 21 | Combine:  T - m 2  a 2  = m 1  a 1   T = m 1  a 1 + m 2  a 2  = (m 1 +m 2 ) a ,[object Object],x y W 1 N 2 N T f 12 T f 21 N 1 W 2
Relative Velocity (review) ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],Preflight 7.1  t = d / v y   Ann v y  = v cos(  ) Beth v y  = v  Carly v y  = v cos(  ) correct A   B   C x y
[object Object],[object Object],ACT  t = d / v y   Ann v y  = v cos(  ) Beth v y  = v  Carly v y  = v cos(  ) A   B   C  x y
[object Object],Swimmer Example  V Ann,ground  = V ann,water +V water,ground x-direction 0 = V x,Ann,Water  + 3 0 = -V Ann,Water  sin(  ) + 3 5 sin(  ) = 3 sin(  ) = 3/5  A   B   C x y 
A   B   C Think of a swimming pool on a cruise ship When swimming to the other side of the pool,  you don’t worry about the motion of the ship ! Demo - bulldozer HW: swimmer x y
Summary of Concepts ,[object Object],[object Object],[object Object],[object Object],[object Object]

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Lecture07

  • 1.
  • 3.
  • 4.
  • 5.
  • 6. Pulley, Incline and 2 blocks N A block of mass m 1 = 2.6 kg rests upon a frictionless incline as shown and is connected to mass m 1 via a flexible cord over an ideal pulley. What is the acceleration of block m 1 if m 2 = 2.0 kg? X – direction  F x = m a x : Block 1: T – m 1 g sin(30) = m 1 a 1x T = m 1 g sin(30) + m 1 a 1x Y – direction F y = m a y : Block 2: T – m 2 g = m 2 a 2y Note: a 1x = - a 2y Combine T – m 2 g = m 2 a 2y m 1 g sin(30) + m 1 a 1x – m 2 g = m 2 a 2y m 1 g sin(30) + m 1 a 1x – m 2 g = -m 2 a 1x m 1 a 1x +m 2 a 1x = m 2 g - m 1 g sin(30) (m 1 +m 2 ) a 1x = g (m 2 - m 1 sin(30)) 1.49 m/s 2 T m 2 g m 1 g T y x y x
  • 7.
  • 8.
  • 9.
  • 10.
  • 11.
  • 12.
  • 13.
  • 14. A B C Think of a swimming pool on a cruise ship When swimming to the other side of the pool, you don’t worry about the motion of the ship ! Demo - bulldozer HW: swimmer x y
  • 15.

Notas do Editor

  1. Transparency?
  2. Do we do example with student pushing me, and show that they go backwards?
  3. Transparency?
  4. Add Calculation. What angle to get straight across river?
  5. Add Calculation. What angle to get straight across river?
  6. Add Calculation. What angle to get straight across river? 36.8 degrees