Question

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?

Short answer

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

Step-by-step solution

Introduction

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.

Explanation 

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 $U_s^{\text{'}}$'.

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

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

Explanation

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 $0.71\mathrm{cm}$.