Two springs of spring constants 1000 N m-1 and 2000 N m-1 are stretched with same force. They will have potential energy in the ratio of

Question

Two springs of spring constants 1000 N m-1 and 2000 N m-1 are stretched with same force. They will have potential energy in the ratio of (A) 2 :1 (B) 22: 12 (C) 1: 2 (D) 12: 22

Tardigrade Admin · Accepted Answer

U=(1/2) k x2 =(1/2) k((F/k))2 =(F2/2 k) U1 × k1=U2 × k2 or (U1/U2)=(k2/k1) =(2000/1000)=(2/1)

Q. Two springs of spring constants $1000 N m^{- 1}$ and $2000 N m^{- 1}$ are stretched with same force. They will have potential energy in the ratio of

$2 : 1$

$2^{2} : 1^{2}$

$1 : 2$

$1^{2} : 2^{2}$

$U = \frac{1}{2} k x^{2}$
$= \frac{1}{2} k (\frac{F}{k})^{2}$
$= \frac{F ^{2}}{2 k}$
$U_{1} \times k_{1} = U_{2} \times k_{2}$
or $\frac{U _{1}}{U _{2}} = \frac{k _{2}}{k _{1}}$
$= \frac{2000}{1000} = \frac{2}{1}$

Q. Two springs of spring constants 1000Nm−1 and 2000Nm−1 are stretched with same force. They will have potential energy in the ratio of

2:1

22:12

1:2

12:22