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 :

Solution

Physics sir · 4 steps

At rest, each spring force equals the block's weight, so write the stored energy in terms of force.

1. Equilibrium stretch. Each block hangs at rest, so the spring pulls up exactly as hard as gravity pulls down: kx = mg. That gives x = mg/k.
2. Energy from force. Put this x into U = 1/2kx². One k cancels, so the stored energy is the weight squared divided by 2k.

   U = 1/2k(mg/k)² = (mg)²/2k ⇒ U_B/U_A = (m_B/m_A)² k_A/k_B

3. Put in numbers. The mass ratio is 200/100 = 2, so its square is 4. From 4k_A = 3k_B, the ratio k_A/k_B = 3/4. So U_B/U_A = 4 × 3/4 = 3.
4. So. The energy stored in B is 3E, which is option (d).

The idea to keep. For a spring holding a hanging mass at rest, use U = F²/2k with F = mg. The extensions are not equal, so do not compare the springs with 1/2kx² at the same x.

Answer from the NTA final answer key. Solution by PrepWiser.