capa9 - 1) Draw free body diagrams: F y N T Ffrict m1g T...

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1) Draw free body diagrams: The net force on m 1 in y-direction is zero: N+F·sin θ - m 1 g=0 b N = m 1 g - F·sin θ b F frict = μ K (m 1 g - F·sin θ ) The net-force in x-direction is: F·cos θ μ K (m 1 g - F·sin θ ) – T = m 1 a From the free body diagram for m 2 we find: T – m 2 g = m 2 a b T = m 2 (a+g) Substituting T in the eqn. for the net force on m 1 yields: a = [F·cos θ μ K (m 1 g - F·sin θ ) – m 2 g]/(m 1 +m 2 ) 2) The spring is mass-less. Therefore it has no momentum (p = m·v = 0·v = 0) The total initial momentum is zero. Therefore, the total momentum will be zero at any time (no external forces.) Use conservation of momentum to determine the speed of each block. The light block will accelerate faster and end up with a larger kinetic energy than the heavy block. The ‘height’ to which each of the masses can climb is proportional to the maximum kinetic energy of each block (remember KE initial +PE initial =KE final +PE final ) 3) This is – of course – another example for conservation of momentum. Here, we have to be careful to understand that if the earth is not part of the system, the force mg acting along the –y direction is an external force. Therefore, the component of the momentum
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capa9 - 1) Draw free body diagrams: F y N T Ffrict m1g T...

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