Two blocks with masses 100 g and 200 g are attached to the ends of springs $A$ and $B$ as shown in f

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 $4 k_A=3 k_B$, is :

Professor PC Thomas Classes · Answer

Assuming the springs are hanging vertically supporting the blocks, the restoring force in each spring at equilibrium equals the weight of its block. $F_A = m_A g = 0.1g$ and $F_B = m_B g = 0.2g$, so $F_B = 2F_A$. The energy stored in a spring is $U = \frac{F^2}{2k}$. For spring A, $E_A = \frac{F_A^2}{2k_A} = E$. For spring B, $E_B = \frac{F_B^2}{2k_B} = \frac{(2F_A)^2}{2k_B} = \frac{4F_A^2}{2k_B} = 4E_A (\frac{k_A}{k_B})$. Given $4k_A = 3k_B \implies \frac{k_A}{k_B} = \frac{3}{4}$. Substituting this gives $E_B = 4E(\frac{3}{4}) = 3E$.