Question

A stretched spring stores 2.0 J of energy. How much energy will be stored if the spring is stretched three times as far?

Short answer

As a result, the spring's new vitality is released$18.0\mathrm{J}$

Step-by-step solution

Introduction

You must exert effort in order to compress or stretch a spring. You must apply a force to the spring that is equivalent to the force the spring applies to you, but in the opposite direction.

Explanation

A spring's potential energy is determined by

$E=\frac{1}{2}k{x}^2$

we have, $k$is the spring constant (a measurement of the spring's stiffness) in $\mathrm{N}/\mathrm{m}$, and$x$is the length in metres that the spring has been stretched beyond its equilibrium point.

Explanation 

The square of the extension is precisely proportional to the spring's potential energy. The relationship between potential energy and extension is defined as follows:

$E\propto x^2$

The following is a representation of the above equation:

$\frac{E^{\text{'}}}{E}=\frac{x^{\text{'}2}}{x^2}$

from above , $E^{\text{'}}$= new energy stored in the spring,$E$is the initial energy stored in the spring, $x$is the starting extension, and $x^{\text{'}}$is the new extension.

all rearrange now$\frac{E^{\text{'}}}{E}=\frac{x^{t2}}{x^2}$as follows:

$E^{\text{'}}=\left(\frac{x^{\text{'}2}}{x^2}\right)E$

Putting $3x$for $x^{\text{'}}$and $2.0\mathrm{J}$for $E$in equation $E^{\text{'}}=\left(\frac{x^{\text{'}2}}{x^2}\right)E$as follows:

$E^{\text{'}}=\left(\frac{(3x{)}^2}{x^2}\right)(2.0\mathrm{J})$

$=18.0\mathrm{J}$

As a result, the spring's new vitality is released $18.0\mathrm{J}$.