Quiz6-Key-Rubric

Quiz6-Key-Rubric - pulling on M to the right is the same as...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Sum of Forces = ma [ ] , F net = ma [ ] , Forces = ma [ ] , F 1 on 2 = F 2 on 1 [ ] , d = x f x i [ ] , v = d t = x f x i t , a avg = Δ v t F g = G Mm R 2 , G = 6.7 × 10 11 [ ] , F E = k Qq R 2 , k = 8.99 × 10 9 [ ] , F k = μ k N [ ] , F s ,max = s N [ ] , F spring = kx [ ] Rubric Code: a) Pulleys can modify either the direction or magnitude of a force. The rules of pulleys are that (I) The tension T is the same throughout the rope and (II) The length of the rope is constant. What are the implications of these rules on the forces on the masses and the motion of the objects for the example below? Explain in 1 sentence. Fall 2010 7B Quiz 6 – Force Model Student ID:________________ Name:________________ , ________________ |______| (last) (first) DL Sec# b) Think about the resulting motion for the system above if there is no friction. Write equations for Newton’s second law for each block. Be careful of which mass you use for each block in F net =ma! c) Solve for T and a using the equations above. M=1kg m= ½kg The implications are that (I) The tension T
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: pulling on M to the right is the same as the tension T pulling on m up and (II) that they have the same acceleration or they move together, or they move at the same rate, etc. T T mg a a Using the above implications we can not write the two equations for M and m that we need. We only need the equation for M in the x-direction and the equation for m in the y-direction. The other equations give us no information. We chose right and down as the positive directions. We then have: F net on M = Ma T = Ma F net on m = ma mg − T = ma mg − Ma = ma ma + Ma = mg a ( m + M ) = mg a = m m + M g = 0.5 1 + 0.5 g = g 3 = 3.27 m s 2 T = Ma = (1)(3.27) = 3.27N Substituting T=Ma into equation 2, we get: There’s a shortcut to skip straight to this step! Can you think of it? It will save us all the work from parts b and c. Part a) 33% Q W E R T Part b) 33% Q W E R T Part c) 33% Q W E R T...
View Full Document

This note was uploaded on 09/08/2010 for the course PHY 7B PHY7B taught by Professor Cebra during the Winter '10 term at UC Davis.

Page1 / 2

Quiz6-Key-Rubric - pulling on M to the right is the same as...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online