Two springs A and B(k_A = 2k_B) are stretched by applying forces of equal magnitudes a the four ends. If the energy stored in A is E, that in B is_A
(a) E/2
(b) 2E
(c) E
(d) E/4

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Solution

(b) 2E

Let x_A and x_B be the extensions produced in springs A and B, respectively.
Restoring force on spring A, $F = k_{A} x_{A}$ ...(i)
Restoring force on spring B, $F = k_{B} x_{B}$ ...(ii)

From (i) and (ii), we get:
$k_{A} x_{A} = k_{B} x_{B}$

It is given that k_A = 2k_B
_{$∴ x_{B} = 2 x_{A}$}

Energy stored in spring A:
$E = \frac{1}{2} k_{A} {x_{A}}^{2}$ ...(iii)

Energy stored in spring B:
$E' = \frac{1}{2} k_{B} {x_{B}}^{2} = \frac{1}{2} (\frac{k_{A}}{2}) (2 x_{A})^{2} ∴ E' = 2 \times (\frac{1}{2} k_{A} {x_{A}}^{2}) = 2 E [From (iii)]$

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Two springs A and B(kA = 2kB) are stretched by applying forces of equal magnitudes a the four ends. If the energy stored in A is E, that in B isA (a) E/2 (b) 2E (c) E (d) E/4

Two springs A and B(k_A = 2k_B) are stretched by applying forces of equal magnitudes a the four ends. If the energy stored in A is E, that in B is_A
(a) E/2
(b) 2E
(c) E
(d) E/4