# Suppose that 6 J of work is needed to stretch a spring from its natural length of 26 cm to a...

## Question:

Suppose that 6 J of work is needed to stretch a spring from its natural length of 26 cm to a length of 47 cm.

(a) How much work is needed to stretch the spring from 34 cm to 39 cm? (Round your answer to two decimal places.)

1.43 J

(b) How far beyond its natural length will a force of 45 N keep the spring stretched? (Round your answer one decimal place.)

_____ cm

## Energy Stored in a Spring:

We use elongation (or stretching) in the spring and the spring constant to evaluate the energy stored in the spring (after stretching or compressing the spring). The below formula shows the energy stored in the spring.

{eq}{E_{spring}} = \dfrac{1}{2}k{x^2} {/eq}

• Here, the spring constant is {eq}k {/eq}
• The stretching in the spring is {eq}x {/eq}

Given Data:

• The work needed to stretch the spring is: {eq}W = 6\;{\rm{J}} {/eq}
• The natural length of the spring is: {eq}{l_0} = 26\;{\rm{cm}} {/eq}
• The final length of the spring is: {eq}l = 47\;{\rm{cm}} {/eq}

(a)

• The spring is stretched from {eq}34\;{\rm{cm}} {/eq} to {eq}39\;{\rm{cm}} {/eq}

Write the expression for the elongation in the spring.

{eq}x = l - {l_0} {/eq}

Substitute the known values.

{eq}\begin{align*} x &= 47\;{\rm{cm}} - 26\;{\rm{cm}}\\[0.3cm] &= 21\;{\rm{cm}} \end{align*} {/eq}

Write the expression for the work needed to stretch the spring.

{eq}W= \dfrac{1}{2}k{x^2} {/eq}

Substitute the known values.

{eq}\begin{align*} 6\;{\rm{J}} &= \dfrac{1}{2}k{\left( {21\;{\rm{cm}}} \right)^2}\\[0.3cm] k &= \dfrac{{2\left( {6\;{\rm{J}}} \right)}}{{{{\left( {21\;{\rm{cm}}} \right)}^2}}}\\[0.3cm] k &= \dfrac{{2\left( {6\;{\rm{J}}} \right)}}{{{{\left( {21\;{\rm{cm}}\left( {\dfrac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)}^2}}}\\[0.3cm] k &= \dfrac{{2\left( {6\;{\rm{J}}} \right)}}{{0.0441\;{{\rm{m}}^2}}}\\[0.3cm] k &= \dfrac{{120000}}{{441}}\;{\rm{N/m}} \end{align*} {/eq}

Write the expression for work needed to stretch the spring from {eq}34\;{\rm{cm}} {/eq} to {eq}39\;{\rm{cm}}. {/eq}

{eq}W = \dfrac{1}{2}k{\left( {39\;{\rm{cm}} - 34\;{\rm{cm}}} \right)^2} {/eq}

Substitute the value of the spring constant in the above equation.

{eq}\begin{align*} W &= \dfrac{1}{2}\left( {\dfrac{{120000}}{{441}}\;{\rm{N/m}}} \right){\left( {5\;{\rm{cm}}} \right)^2}\\[0.3cm] &= \dfrac{1}{2}\left( {\dfrac{{120000}}{{441}}\;{\rm{N/m}}} \right){\left( {5\;{\rm{cm}}\left( {\dfrac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)^2}\\[0.3cm] &= 0.340136\;{\rm{J}}\\[0.3cm] &\approx 0.34\;{\rm{J}} \end{align*} {/eq}

Thus, it requires {eq}\boxed{\color{blue}{0.34\;{\rm{J}}}} {/eq} to stretch the spring.

(b)

• The spring force applied to the spring is: {eq}F = 45\;{\rm{N}} {/eq}

Write the expression for the spring force.

{eq}F = kx {/eq}

Here, the stretching in the spring beyond its natural length is {eq}x. {/eq}

Substitute the known values.

{eq}\begin{align*} 45\;{\rm{N}} &= \left( {\dfrac{{120000}}{{441}}\;{\rm{N/m}}} \right)x\\[0.3cm] x &= \dfrac{{45\;{\rm{N}}}}{{\dfrac{{120000}}{{441}}\;{\rm{N/m}}}}\\[0.3cm] x &= 0.165375\;{\rm{m}}\\[0.3cm] x &= 0.165375\;{\rm{m}}\left( {\dfrac{{100\;{\rm{cm}}}}{{1\;{\rm{m}}}}} \right)\\[0.3cm] x &= 16.5375\;{\rm{cm}}\\[0.3cm] x &\approx 16.5\;{\rm{cm}} \end{align*} {/eq}

Thus, the given amount of force will stretch the spring by {eq}\boxed{\color{blue}{16.5\;{\rm{cm}}}} {/eq}.

Practice Applying Spring Constant Formulas

from

Chapter 17 / Lesson 11
3.4K

In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.