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A block `A` of mass `m` is over a plank `B` of mass `2m`. Plank `B` si placed over a smooth horizontal surface. The coefficient of friction between `A` and `B` is `(1)/(2)`. Blocks `A` is given a velocity `V_(0)` towards right. Then A. Acceleration of `A` is `(g)/(2)`B. Acceleration of `A` is `g`C. Acceleration of `B` relative to `A` is `(3)/(4)g`D. Acceleration of `A` is zero |
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Answer» Correct Answer - A::C Force of friction between the two will be maximum i.e., `mu mg` Retardation of `A` is `a_(A) = (mu mg)/(m) = mu g` and acceleration of `B` is `a_(B) = (mu mg)/(2m) = (mu g)/(2)` `therefore` Acceleration of `B` relative to `A` is `a_(BA) = a_(A)+a_(B) = (3 mu g)/(2)` substituting `mu = (1)/(2)a_(BA) = (3g)/(4)`. |
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