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A block of mass `m = 2 kg` is connected to a spring of force constant `k = 50 n//m`. Initially the block is at rest and the spring has natural length . A constant force `f = 60 N`, is applied horizontally towards right, the maximum speed of the block (in `m//s` will be (neglect frication). |
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Answer» Correct Answer - 6 For maximum speed. `(dv)/(dt) = 0 rArr a = 0` It at maximum speed, elongation is `x`, then `F = kx rArr x = (F)/(k)` Now, by `WET`. `-(1)/(2)kx^(2) + Fx = (1)/(2) mv^(2)` `V = (F)/(sqrt(mk)) = 6 m//s` |
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