Two springs A and B having spring constant KA and KB(KA=2 KB) are stretched by applying force of equal magnitude. If energy stored in spring A is EA then energy stored in B will be

Question

Two springs A and B having spring constant KA and KB(KA=2 KB) are stretched by applying force of equal magnitude. If energy stored in spring A is EA then energy stored in B will be (A) 2EA (B) EA/4 (C) EA/2 (D) 4EA

Tardigrade Admin · Accepted Answer

Energy =(1/2) K x2=(1/2) (F2/K) . (KA/KB)=2 ∴ (EA/EB)=(1/2) . or EB=2 EA .

Q. Two springs $A$ and $B$ having spring constant $K_{A}$ and $K_{B} (K_{A} = 2 K_{B})$ are stretched by applying force of equal magnitude. If energy stored in spring $A$ is $E_{A}$ then energy stored in $B$ will be

$2 E_{A}$

$E_{A} /4$

$E_{A} /2$

$4 E_{A}$

Energy $= \frac{1}{2} K x^{2} = \frac{1}{2} \frac{F ^{2}}{K} .$
$\frac{K _{A}}{K _{B}} = 2$
$∴ \frac{E _{A}}{E _{B}} = \frac{1}{2} .$
or $E_{B} = 2 E_{A} .$

Q. Two springs A and B having spring constant KA​ and KB​(KA​=2KB​) are stretched by applying force of equal magnitude. If energy stored in spring A is EA​ then energy stored in B will be

2EA​

EA​/4

EA​/2

4EA​