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(B) For the lower plate of mass $m$ to lift off the ground,the upward spring force must be equal to its weight.Let $k$ be the spring constant. The con

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(B) For the lower plate of mass $m$ to lift off the ground,the upward spring force must be equal to its weight.Let $k$ be the spring constant. The condition for the lower plate to lift is $kx = mg$,where $x$ is the extension of the spring from its natural length.So,$x = \frac{mg}{k}$.Now,consider the energy conservation between the initial compressed state (position $I$) and the final extended state (position $II$) where the lower plate just lifts off.Let $h$ be the initial compression of the spring when weight $W$ is placed on the upper plate.The total energy at position $I$ is the potential energy of the compressed spring: $U_I = \frac{1}{2}kh^2$.The total energy at position $II$ is the potential energy of the extended spring plus the potential energy of the upper plate of mass $m$: $U_{II} = \frac{1}{2}kx^2 + mgh_{total}$,where $h_{total} = h + x$ is the total height change of the upper plate.By energy conservation: $\frac{1}{2}kh^2 = mg(h+x) + \frac{1}{2}kx^2$.Substituting $x = \frac{mg}{k}$:$\frac{1}{2}kh^2 = mgh + \frac{m^2g^2}{k} + \frac{1}{2}k(\frac{mg}{k})^2 = mgh + \frac{m^2g^2}{k} + \frac{m^2g^2}{2k} = mgh + \frac{3m^2g^2}{2k}$.Multiplying by $2/k$: $h^2 - \frac{2mgh}{k} - \frac{3m^2g^2}{k^2} = 0$.Solving for $h$: $h = \frac{\frac{2mg}{k} + \sqrt{(\frac{2mg}{k})^2 + 4(\frac{3m^2g^2}{k^2})}}{2} = \frac{\frac{2mg}{k} + \sqrt{\frac{16m^2g^2}{k^2}}}{2} = \frac{\frac{2mg}{k} + \frac{4mg}{k}}{2} = \frac{3mg}{k}$.At equilibrium in position $I$,the spring force balances the weight of the upper plate and the weight $W$: $kh = mg + W$.Substituting $h = \frac{3mg}{k}$: $k(\frac{3mg}{k}) = mg + W \Rightarrow 3mg = mg + W \Rightarrow W = 2mg$.Thus,the weight $W$ must be just more than $2mg$.

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