Two springs A and B have constants Kₐ=2 $K_{b}$ . Equal forces stretch them. If energy in A is Eₐ, energy stored in B is:
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Question

Two springs A and B have constants Kₐ=2 K b K_{b} K b ​ . Equal forces stretch them. If energy in A is Eₐ, energy stored in B is:

G1 Physics · Accepted Answer

Step 1 — Principle: Elastic spring potential energy is Uₛ = ½kx². Step 2 — Working: For equal applied force, spring energy is U=F²/(2k), hence Eᵦ/Eₐ=Kₐ/Kᵦ=2. Step 3 — Working: Thus Eᵦ=2Eₐ. Correct answer: A — 2Eₐ.

Two springs A and B have constants Kₐ=2 K b K_{b} K b ​ . Equal forces stretch them. If energy in A is Eₐ, energy stored in B is:

Two springs A and B have constants Kₐ=2 K b K_{b} K b . Equal forces stretch them. If energy in A is Eₐ, energy stored in B is: