Question

How far must you compress a spring with twice the spring constant to store the same amount of energy?

Answer 1

The spring with twice spring constant has to compress $\boxed{\frac{1}{{\sqrt 2 }}}$ or $\boxed{0.707}$ times the compression of spring of the given spring constant.

Further explanation:

Consider a spring of spring constant k is compressed by length $x$.

Then amount of energy stored in the spring 1 can be calculated as.

${E_1}=\dfrac{1}{2}k{x^2}$                                                                 …… (1)

Here, ${E_1}$ is the amount of energy store in spring 1 due to its compression.

Now, another spring with spring constant $2k$ as given is compressed by length $y$ .

Then the energy stored in the spring due to its compression can be calculated as.

${E_2}=\dfrac{1}{2}\left( {2k}\right){y^2}$                            …… (2)

Here, ${E_2}$ is the amount of energy store in spring 2 due to its compression.

As given in the question energy store in both the spring is same. So, equate equation (1) and equation (2).

$\begin{aligned}\frac{1}{2}k{x^2}&=\frac{1}{2}\left( {2k} \right){y^2}\\k{x^2}&=\left( {2k}\right){y^2}\\\frac{1}{2}{x^2}&={y^2}\\\end{aligned}$

   

Taking square root both the sides in above equation,

$\begin{gathered}\sqrt {\frac{1}{2}{x^2}}=\sqrt {{y^2}}\\\frac{x}{{\sqrt 2 }} = y \\ \end{gathered}$

 

The above equation can be written as,

$\boxed{y=\frac{x}{{\sqrt 2 }}}$

 

So, spring of spring constant $2k$ is compressed $\dfrac{1}{{\sqrt 2 }}$ times the compression of spring of spring constant $k$.

Learn more:

1. Conservation of energy: brainly.com/question/3943029.

2. Motion under friction brainly.com/question/7031524.

3. Conservation of momentum brainly.com/question/9484203.

Answer detail:

Grade: Senior school

Chapter: Work and Energy

Subject: Physics

Keywords:

Compression, expansion, spring, twice the spring constant, same amount of energy, spring constant, energy, potential energy, kinetic energy, force, Hook’s law.

Answer 2

Explanation :

The energy stored in the spring is called as elastic potential energy.

Mathematically, it is written as

$U=\dfrac{1}{2}kx^2$

Where,

k is the spring constant

x is the displacement from mean position.

Initially,

$U_1=\dfrac{1}{2}k_1x_1^2$

and

$U_2=\dfrac{1}{2}k_1x_2^2$

Since, it store the same amount of energy.

$\dfrac{1}{2}k_1x_1^2=\dfrac{1}{2}k_1x_2^2$

$k_2=2k_1$

$x_1^2=2x_2^2$

$x_2=\dfrac{x_1}{\sqrt{2} }$

$x_2=0.707\ x_1$

So, a spring with twice the spring constant compresses 0.707 times.

Hence, this is the required solution.