# If it requires 7.0J of work to stretch a particular spring by 2.3 cm from its equilibrium length,...

## Question:

If it requires 7.0J of work to stretch a particular spring by 2.3 cm from its equilibrium length, how much more work will be required to stretch it an additional 3.5 cm?

## Work:

In physics, the physical quantity work is defined as the product of applied force on an object and the object's total displacement due to force. It is a scalar quantity, and it can be positive, negative, or zero depends on the applied force and the distance covered.

## Answer and Explanation: 1

**Given data**

- The initial length of the stretch from equilibrium position is: {eq}{x_1} = 2.3\;{\rm{cm}} {/eq}

- The amount of work done to stretch the spring to the length {eq}{x_1} {/eq} is: {eq}{W_1} = 7.0\;{\rm{J}} {/eq}

- The additional length of the spring that is required to stretch is: {eq}{x_2} = 3.5\;{\rm{cm}} {/eq}

As, there is no loss of energy, so by using energy conservation the potential energy stored in the spring due to the stretch will be equal to the work done required to stretch the spring.

The expression for the work required to stretch the spring to the distance {eq}{x_1} {/eq} from equilibrium is given as follows,

{eq}{W_1} = \dfrac{1}{2}kx_1^2 {/eq}

Here, {eq}k {/eq} is the spring constant.

Substitute all the values in the above expression,

{eq}\begin{align*} 7\;{\rm{J}} &= \dfrac{1}{2} \times k \times {\left( {2.3\;{\rm{cm}} \times \dfrac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)^2}\\ k &= 26465\;{{\rm{J}} {\left/ {\vphantom {{\rm{J}} {\rm{m}}}} \right. } {\rm{m}}} \end{align*} {/eq}

Calculate the total length of stretch from equilibrium position,

{eq}\begin{align*} x &= {x_1} + {x_2}\\ x &= 2.3\;{\rm{cm}} + {\rm{3}}{\rm{.5}}\;{\rm{cm}}\\ x &= 5.8\;{\rm{cm}} \end{align*} {/eq}

The expression for the work required to stretch the spring to the length {eq}x {/eq} from the equilibrium is given as follows,

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

Substitute all the values in the above expression,

{eq}\begin{align*} W &= \dfrac{1}{2} \times 26465 \times {\left( {5.8\;{\rm{cm}} \times \dfrac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)^2}\\ W &= 44.5\;{\rm{J}} \end{align*} {/eq}

Calculate the additional work required to stretch the spring from length {eq}{x_1} {/eq} to the length {eq}x {/eq} is given as follows,

{eq}\begin{align*} {W_2} &= W - {W_1}\\ {W_2} &= 44.5\;{\rm{J}} - 7.0\;{\rm{J}}\\ {W_2} &= 37.5\;{\rm{J}} \end{align*} {/eq}

Thus, the additional work required to stretch the spring is {eq}37.5\;{\rm{J}} {/eq}.

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from

Chapter 17 / Lesson 11In 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.