A spring is compressed $1.0cm$. How far must you compress a spring with twice the spring constant to store the same amount of energy?

The units for the spring constant are Newton per metre. We'll look at the spring constant definition now that we've established that k is the spring constant. The stiffness of the spring is defined as the spring constant.

For the spring, the fresh spring energy is,

$U_{\mathrm{s}}{}^{\text{'}}=\frac{1}{2}{k}^{\text{'}}{\left(\Delta s^{\text{'}}\right)}^2$

Here, $k^{\text{'}}$is the new spring constant of the spring and $\Delta s^{\text{'}}$is the compression of the spring. The new spring constant of the spring is,

$k^{*}=2k$

Substitute $2k$for $k^{\text{'}}$in the equation $U_s^{\text{'}}=\frac{1}{2}{k}^{\text{'}}{\left(\Delta s^{\text{'}}\right)}^2$and solve for $U_s^{\text{'}}$'.

$U_s^{\text{'}}=\frac{1}{2}\left(2k\right){\left(\Delta s^{\text{'}}\right)}^2$

$=k{\left(\Delta s^{\text{'}}\right)}^2$

Given that a spring with twice the spring constant stores the same amount of energy as a spring with half the spring constant.

$U{}^{\text{'}}{}^{\text{'}}=U_{\mathrm{s}}$

Substitute $\frac{1}{2}k(1.0\mathrm{cm}{)}^2$for $U_{\mathrm{s}}$and $k{\left(\Delta s^{\text{'}}\right)}^2$for $U_{\mathrm{s}}^{\text{'}}$.

$k{\left(\Delta s^{\text{'}}\right)}^2=\frac{1}{2}k(1.0\mathrm{cm}{)}^2$

Now, solve for $\Delta s$.

${\left(\Delta s^{\text{'}}\right)}^2=\frac{1}{2}(1.0\mathrm{cm}{)}^2$

$\Delta s^{\text{'}}=\sqrt{0.5{\mathrm{cm}}^2}$

$=0.707\mathrm{cm}$

Round off to two significant figures, the spring must be compressed to $0.71\mathrm{cm}$.