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Question

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


A

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B

2E

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C

E

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D

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Solution

The correct option is B

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)


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