# Question #4eb4a

Sep 30, 2017

Think it to be $\frac{E}{2}$
Hope it helps...

#### Explanation:

Suppose,the spring $A$ and $B$ stretches by the length ${x}_{1}$ and ${x}_{2}$ unit.
We know that Increase in length is directly proportional to force applied but negative in sign.
So we can write,
$F = - k {x}_{1}$ and $F = - 2 k {x}_{2}$ [because each time force used is equal]
From the equation we can write that $- k {x}_{1} = - 2 k {x}_{2} \implies {x}_{1} = 2 {x}_{2}$
Here,we also know for the spring $A$ ,
$E = \frac{1}{2} k {x}_{1}^{2}$
So for the spring $B$ the energy restored will be $\frac{1}{2} \times 2 k \times {x}_{2}^{2} \implies \frac{1}{2} \times 2 k \times {x}_{1}^{2} / 4 = \frac{E}{2}$