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Unformatted text preview: Physics 140  Discussion 07  Solutions 5.34. I DENTIFY : Constant speed means zero acceleration for each block. If the block is moving, the friction force the tabletop exerts on it is kinetic friction. Apply m Σ = F a r r to each block. S ET U P : The freebody diagrams and choice of coordinates for each block are given by Figure 5.34. 4.59 kg A m = and 2.55 kg. B m = E XECUTE : (a) y y F ma Σ = with y a = applied to block B gives B m g T = and 25.0 N. T = x x F ma Σ = with x a = applied to block A gives k T f = and k 25.0 N. f = 45.0 N A A n m g = = and k k 25.0 N 0.556. 45.0 N A f n μ = = = (b) Now let A be block A plus the cat, so 9.18 kg. A m = 90.0 N A n = and k k (0.556)(90.0 N) 50.0 N. f n μ = = = x x F ma ∑ = for A gives k . A x T f m a = y Fy ma ∑ = for block B gives . B B y m g T m a = x a for A equals y a for B , so adding the two equations gives k ( ) B A B y m g f m m a = + and 2 k 25.0 N 50.0 N 2.13 m/s . 9.18 kg 2.55 kg B y A B m g f a m m = = =  + + The acceleration is upward and block B slows down. E VALUATE : The equation k ( ) B A B y m g f m m a = + has a simple interpretation. If both blocks are considered together then there are two external forces: B m g that acts to move the system one way and k f that acts oppositely. The net force of k B m g f must accelerate a total mass of . A B m m + Figure 5.34 __________________________________________________________________________________ 5.36. I DENTIFY : Apply m Σ = F a r r to the box. When the box is ready to slip the static friction force to the box....
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 Winter '10
 Uher
 Acceleration, Force, Friction

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