Laws of MotionHard
Question
Two blocks with masses 100 g and 200 g are attached to the ends of springs A and B as shown in figure. The energy stored in A is E . The energy stored in B , when spring constants $k_{A},k_{B}$ of A and $B$, respectively satisfy the relation $4k_{A} = 3k_{B}$, is :
Options
A.4E
B.2E
C.3E
D.$\frac{4}{3}E$
Solution
For equilibrium $kx = mg$
$${U = \frac{1}{2}{kx}^{2} = \frac{1}{2}\frac{{\text{ }m}^{2}{\text{ }g}^{2}}{k} }{U \propto \frac{m^{2}}{k} }{\frac{U_{A}}{U_{B}} = \left( \frac{m_{A}}{m_{B}} \right)^{2}\frac{k_{B}}{k_{A}} = \left( \frac{1}{2} \right)^{2}\left( \frac{4}{3} \right) = \frac{1}{3} }{\frac{E}{U_{B}} = \frac{1}{3} \Rightarrow U_{B} = 3E}$$
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