ps06sol - 8.01T Problem Set 6 Solutions Fall 2004 Problem...

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8.01T Problem Set 6 Solutions Fall 2004 Problem 1. Experiment 5 b) Analysis. a) We know that when the string is about to slide, the tension varies as T ( ) = T 0 e µ s . It means that T 2 = T 1 e µ s ± . On the other hand, analyzing the free-body diagrams for the masses, we conclude that T 1 = m 1 g and T 2 = m 2 g . Thus the maximum ratio of the masses before the string starts to slide is m 1 = e µ s ± m 2 b) If we assume that the string is massless, the derivation of the tension of the string, which is given in the description of Experiment 5 b) is unchanged, except that the friction force is now µ k N . Thus the tension varies as T ( ) = T 0 e µ k . Since both masses move with the same acceleration, T 1 m 1 g = m 1 a and T 2 m 2 g = m 2 a . Thus m 2 ( g + a ) = m 1 ( g a ) e µ k ± . The solution of this equation is m 1 e µ k ± m 2 a = g m 2 + m 1 e µ k ± Problem 2. Experiment 6 Pre-Lab Question. a) The work done on the cart by the force of gravity before is comes in contact with the spring is W = ( mg sin ) l
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ps06sol - 8.01T Problem Set 6 Solutions Fall 2004 Problem...

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