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Block A has a weight of `3000 N` and block B has a weight of `500 N`. Determine the distance that A must descend from rest before it obtains a speed of `2.5 m//s`. Neglect the mass of the cord and pulleys. . |
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Answer» If A descends X then B will ascend 2x. Further if speed of A at this instant is `2.5 m//s`, then speed of B at this instant will be `5 m//s`. Now, Deecrease in potential energy of `A =` increase in potential energy of `B +` increase in kinetic energy of both `:. (300)x =(50)(2x) + 1/2 ((300)/(9.8)) (2.5)^(2)` `+(1)/(2) ((50)/(9.8))(5.0)^(2)` Solving we get, `x=0.796 m`. |
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