Question

Two springs a and b having spring constants Ka and KB with Ka=2kb are stretched by applying a force of equal magnitude .If energy stored in spring A is Ea what is the energy stored in spring B

Answer 1

Answer:

Explanation:

Let the extension produced in first spring be x

1/2 ka x2 = E

F = ka x

Eliminating x from the above two equations

x2 = F2 / ka2

E = 1/2 ka (F2 / ka2) = 1/2 F2 / ka

Energy stored is inversely proportional to k, because 1/2 F2 is constant.

Let energy stored in B be E"

E" = 1/2 F2 / kb

Taking ratios of E" and E

E"/E = ka/ kb = 2

E" = 2E

Thus energy stored in second spring is 2E