Q. Two springs have force constant K1 and K2 ( K1> K2). Each spring is extended by same force F. Ifthe elastic potential energy are E1 and E2 then E1/E2 is K1/K2 K2/K1 ✓(K1/K2) ✓(K2/K1)
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Solution
Expansion of 1st string change in x= f/k1
expansion in 2nd string change in x= f/k2
sso, ΔE1/ΔE2= 1/2 k1(Δx1)² divided by 1/2 k2(Δx2)² E1/E2= k2/k1