InterviewSolution
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InterviewSolution
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Two block `A` and `B` of masses `1kg` and`2kg` respectively are connected by a string, passing over a light frictionless pulley `B` as shown. Another string connect the center of pulley. Both the blocks are resting on a horizontal floor and the pulley is help such that string remains just taut. At moment `t = 0`, a force `F = 20 t` starts acting on the pully along vertically upwards direction as shown in figure.Calculate (a) velicity of `A` when `B` loses contact with the floor. (b) height raised by the pulley upto that instant. (Take =`g =10 m//s^(2))` |
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Answer» (a) Let `T` be the tansion in the string. Then, `2T =20 t` `T =20 t` newton Let the block `A` loses its contact with the floor at time `t =t_(1)`. This happens when the tension in string becomes equal to the weight of `A` . Thus, `T =mg` or `10t_(2) =1xx10` or `t_(2) =1s` ...(i) Similarly, for block `B` we have `10t_(2) =2xx10` or `t_(2) =2s` ...(ii) i.e. the block `B` loses contact at `2s`. For block `A`, at time `t` such that `tget_(1)` let a be its acceleration in upward direction.Then, `10t-1xx10 =1xxa = (dv//dt)` `du =10(t-1)dt` ...(iii) Integrating this expression, we get `int_(0)^(v) dv = 10 int_(1)^(t) (t-1) dt` or `v = 5t^(2)-10t+5` ...(iv) Substituting `t = t_(2) = 2s` `v = 20 - 20 + 5 = 5m//s` ...(v) (b) From Eq. (iv). `dy = (5t_(2)-10t + 5)dt` ...(vi) when, `y` is the vertical displacement of block `A` at time `t (ge t_(1))`. Integrating, we have `int_(y = 0)^(y =h) =int_(t = 1)^(t =2) (5t^(2)-10t + 5)dt` `h = 5[(t^(3))/(3)]_(1)^(2) -10 [(t^(2))/(2)]_(1)^(2) + 5[t]_(1)^(2) = (5)/(3)m` :. Height raised by pulley upto that instant `=(h)/(2) = (5)/(6) m` |
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