Two springs have force constant K1 and K2(K1 K2). Each spring is extended by same force F. It their elastic potential energy are E1 and E2 then (E1/E2) is

Question

Two springs have force constant K1 and K2(K1 > K2). Each spring is extended by same force F. It their elastic potential energy are E1 and E2 then (E1/E2) is (A) (K1/K2) (B) (K2/K1) (C) √(K1/K2) (D) √(K2/K1)

Tardigrade Admin · Accepted Answer

x=(F/K) U=(1/2) K x2=(1/2) K((F/K))2 U=(F2/2 K) U α (1/K) (U1/U2)=(K2/K1)

Q. Two springs have force constant $K_{1}$ and $K_{2} (K_{1} > K_{2})$ . Each spring is extended by same force $F$ . It their elastic potential energy are $E_{1}$ and $E_{2}$ then $\frac{E _{1}}{E _{2}}$ is

$\frac{K _{1}}{K _{2}}$

$\frac{K _{2}}{K _{1}}$

$\frac{K _{1}}{K _{2}}$

$\frac{K _{2}}{K _{1}}$

$x = \frac{F}{K}$
$U = \frac{1}{2} K x^{2} = \frac{1}{2} K (\frac{F}{K})^{2}$
$U = \frac{F ^{2}}{2 K}$
$Uα \frac{1}{K}$
$\frac{U _{1}}{U _{2}} = \frac{K _{2}}{K _{1}}$

Q. Two springs have force constant K1​ and K2​(K1​>K2​). Each spring is extended by same force F. It their elastic potential energy are E1​ and E2​ then E2​E1​​ is

K2​K1​​

K1​K2​​

K2​K1​​​

K1​K2​​​

x=KF​ U=21​Kx2=21​K(KF​)2 U=2KF2​ UαK1​ U2​U1​​=K1​K2​​

Q. Two springs have force constant $K_{1}$ and $K_{2} (K_{1} > K_{2})$ . Each spring is extended by same force $F$ . It their elastic potential energy are $E_{1}$ and $E_{2}$ then $\frac{E _{1}}{E _{2}}$ is

$\frac{K _{1}}{K _{2}}$

$\frac{K _{2}}{K _{1}}$

$\frac{K _{1}}{K _{2}}$

$\frac{K _{2}}{K _{1}}$

$x = \frac{F}{K}$
$U = \frac{1}{2} K x^{2} = \frac{1}{2} K (\frac{F}{K})^{2}$
$U = \frac{F ^{2}}{2 K}$
$Uα \frac{1}{K}$
$\frac{U _{1}}{U _{2}} = \frac{K _{2}}{K _{1}}$