Two blocks with masses 100 text{ g} and 200 t | vedclass

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(C) For a spring-mass system in equilibrium,the spring force balances the gravitational force: $kx = mg$,where $x$ is the extension of the spring.Thus

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$3E$

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(C) For a spring-mass system in equilibrium,the spring force balances the gravitational force: $kx = mg$,where $x$ is the extension of the spring.Thus,the extension is $x = \frac{mg}{k}$.The elastic potential energy $U$ stored in the spring is given by $U = \frac{1}{2}kx^2$.Substituting the value of $x$,we get $U = \frac{1}{2}k\left(\frac{mg}{k}ight)^2 = \frac{m^2g^2}{2k}$.From this expression,we see that $U \propto \frac{m^2}{k}$.Given $m_{A} = 100 	ext{ g}$,$m_{B} = 200 	ext{ g}$,and $4k_{A} = 3k_{B}$,we can write the ratio of energies as:$\frac{U_{A}}{U_{B}} = \left(\frac{m_{A}}{m_{B}}ight)^2 \cdot \left(\frac{k_{B}}{k_{A}}ight)$.Substituting the values: $\frac{E}{U_{B}} = \left(\frac{100}{200}ight)^2 \cdot \left(\frac{4}{3}ight) = \left(\frac{1}{2}ight)^2 \cdot \left(\frac{4}{3}ight) = \frac{1}{4} \cdot \frac{4}{3} = \frac{1}{3}$.Therefore,$U_{B} = 3E$.

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