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(B) The velocity of the block is maximum when its acceleration is zero. At any compression $x$ of the spring,the forces acting on the block are the gr

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$15$

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(B) The velocity of the block is maximum when its acceleration is zero. At any compression $x$ of the spring,the forces acting on the block are the gravitational force $mg$ (downward) and the spring force $kx$ (upward). The net force $F_{net} = mg - kx$. For maximum velocity,the net force must be zero,so $mg - kx = 0$,which gives $x = mg/k$. Given $m = 1 \ kg$,$g = 10 \ m/s^2$ (using $g = 10 \ m/s^2$ for standard calculation),and $k = 2 \ kg/cm = 200 \ N/m$. Converting units: $k = 2 \ kg/cm = 200 \ N/m = 2 \ N/cm$. Thus,$x = (1 \ kg 	imes 10 \ m/s^2) / (200 \ N/m) = 10/200 \ m = 0.05 \ m = 5 \ cm$. The height $h$ from the table surface is the original length of the spring minus the compression $x$. $h = L - x = 20 \ cm - 5 \ cm = 15 \ cm$.

$A$ light spring of length $20 \ cm$ and force constant $2 \ kg/cm$ is placed vertically on a table. $A$ small block of mass $1 \ kg$ falls on it. The height $h$ from the surface of the table at which the block will have the maximum velocity is ............... $cm$.

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