Question

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

Answer 1

Explanation:

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

Answer 2

Option (b) 2 E.

Explanation:

As per as the given conditions, $\mathrm{K}_{\mathrm{A}}=2 \mathrm{K}_{\mathrm{B}}$.

For the constant force on both springs,

Therefore,    

                $\mathrm{m}_{\mathrm{B}}=2 \mathrm{m}_{\mathrm{A}}$

The energy stored in the springs A is,

Thus                      A = E  ------> eqn ( 1 )

                   Also,  E = F × $m_A$

Where, that constant force is F.

The energy stored in the spring B is,

Stored Energy in the B = F ×  $m_B$  -----> eqn ( 2 )

Energy stored in B = 2 E  

Energy stored in B from equations ( 1 ) and ( 2 ) is twice that of A (i.e.,) 2 E.

Hence, option ( b ) 2 E is right answer.