Question

Two springs A and B (k_A = 2k_B) are stretched by applying forces of equal magnitudes at their two ends. If the energy stored in A is E, that in B is:
1. 4E
2. 2E
3. E
4. E / 2

Answer 1

Correct Answer - Option 2 : 2E

Concept:

Potential energy of spring: The energy associated with the state of compression or extension of an elastic spring.

• The spring is compressed or elongated, it tends to recover its original length on account of elasticity.
• For small stretch or compression, it follows Hook’s law.
• Restoring force ∝ stretch or compression, - F ∝ x

- F = k x 

The negative sign represents that restoring force always directed towards the equilibrium position.

• Potential energy of stretched or compressed spring

E = ½ kx2 

Where k = spring constant, x = extension or compression in metre.

Calculation:

Given: K_A = 2 K_B

From Hook's law: Restoring force is given as-

- F = k x      - - - - - (1)

And energy in spring can be given as-

• Potential energy of stretched or compressed spring

E = ½ kx2  - - - - - (2)

Where k = spring constant, x = extension or compression in metre.

From (1) & (2); we get

\(E = \frac{{{F^2}}}{{2\;K\;}} \Rightarrow E \propto \frac{1}{K}\)

Given that:   K_A = 2 K_B

_{\(\frac{{{E_A}}}{{{E_B}}} = \frac{{{K_B}}}{{{K_A}}} = \frac{{{K_B}}}{{2{K_B}}}\)}

∴ K_B = 2 E_A