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A block of mass `m = 2 kg` is connected to a a spring of force constant `k = 50 N//m`. Initially the block is at rest and tlhe spring has natural length. A constant force `F = 60 N` is applied horizontally towards, the maximum speed of the block (in `m//s`) will be (negelect friction). |
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Answer» Correct Answer - 6 For maximum speed, `(dv)/(dt) = 0 rArr a = 0` If 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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