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 :

Step-by-step solution

For blocks hanging in equilibrium, spring extensions are: x_A = m_A g k_A = 100g k_A and x_B = m_B g k_B = 200g k_B . Energy stored: E = 1 2 k_A x_A^2 = 5000g^2 k_A and U_B = 1 2 k_B x_B^2 = 20000g^2 k_B . Taking ratio: U_B E = 4 k_A k_B . Given 4k_A = 3k_B , we have k_A k_B = 3 4 . Therefore: U_B = 4E 3 4 = 3E .