Two blocks with masses 100g and 200g 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 :
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Question

Two blocks with masses 100g and 200g 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 :

Options

G1 Physics · Accepted Answer

For a mass hanging in equilibrium from a vertical spring, kx = mg. Therefore, x = mg/k. Elastic potential energy stored in the spring is U = ½kx² = ½k(mg/k)² = m²g²/(2k). Thus, U ∝ m²/k. For spring A: m(A) = 100 g. For spring B: m(B) = 200 g = 2m(A). Given, 4kA = 3kB, so kB/kA = 4/3. Hence, U(A)/U(B) = [m(A)/m(B)]² × [kB/kA] = (1/2)² × 4/3 = 1/3. Since U(A) = E, we get E/U(B) = 1/3. Therefore, U(B) = 3E. Correct option: C.