Concrete blocks of masses $m_{A}$ and $m_{B}$ are balanced on two identical vertical springs. $m_{A} = 2 m_{B}$ . The gravitational potential energy of each system is zero at the equilibrium position of the springs. Which statement is true for the total mechanical energy of the systems when the blocks are balanced on the springs?

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$E_{A} = 2 E_{B}$

$E_{A} = - 2 E_{B}$

$E_{A} = E_{B}$

$E_{A} = 4 E_{B}$

answer is C.

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Detailed Solution

${Kx}_{A} = m_{A} g$

${Kx}_{B} = m_{B} g \Rightarrow \frac{x_{A}}{x_{B}} = \frac{m_{A}}{m_{B}} = 2 \Rightarrow x_{A} = 2 x_{B}$

${PE}_{A} = \frac{1}{2} {Kx}^{2}_{A} {PE}_{B} = \frac{1}{2} {Kx}_{B}^{2} \Rightarrow \frac{{PE}_{A}}{{PE}_{B}} = {(\frac{x_{A}}{x_{B}})}^{2} = 4$

So for A, potential energy of spring is four times that for B.

So $E_{A} = 4 E_{B}$

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