Two springs A and B (kA = 2kB) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, that in B is
2E
Let the extension in the spring be x. Let's draw the free body drawaing of one end of the spring A (Right End)
KAx
F = KAx,x = FKA
Now we know that the potential energy stored in a streched spring is given by 12k x2
⇒ (12 KAFKA)2
⇒ 12 F2(KA)
Given Energy stored in A= E
⇒ 12 F2(KA) = E ..(1)
kBx
X = FkB
Energy = 12 kB
Similarly , for spring B
X = FkB
Energy = 12 KAFKA2 = 12 F2(KB)
Given kB = kA2)
⇒ Energy = 12 F2(KA) 2
Energy B = 2E .. (ii)