# A hand exerciser utilizes a coiled spring. A force of 105.0 N is required to compress the string...

## Question:

A hand exerciser utilizes a coiled spring. A force of 105.0 N is required to compress the string by 0.0191 m. Determine the force needed to compress the spring by 0.0440 m.

## Spring Force:

The spring force constant is evaluated using the force required to compress the spring and the magnitude of the compression or elongation. The spring force constant is measured in newton per meter, and it depends on the spring's material.

Given Data:

• The magnitude of the force required to stretch the spring is: {eq}{F_1} = 105.0\;{\rm{N}} {/eq}
• The initial compression in the spring is: {eq}{x_1}= 0.0191\;{\rm{m}} {/eq}
• The final compression in the spring is: {eq}{x_2} = 0.0440\;{\rm{m}} {/eq}

Write the expression for the spring force required to stretch the spring (initially).

{eq}{F_1} = k{x_1} {/eq}

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

Substitute the known values.

{eq}\begin{align*} \left( {105.0\;{\rm{N}}} \right) &= k\left( {0.0191\;{\rm{m}}} \right)\\[0.3cm] k &= \dfrac{{105.0\;{\rm{N}}}}{{0.0191\;{\rm{m}}}}\\[0.3cm] k &= \dfrac{{105.0}}{{0.0191\;{\rm{m}}}}\;{\rm{N/m}} \end{align*} {/eq}

Write the expression for the force required to stretch the spring finally.

{eq}{F_2} = k{x_2} {/eq}

Substitute the known values.

{eq}\begin{align*} {F_2} &= \left( {\dfrac{{105.0}}{{0.0191\;{\rm{m}}}}\;{\rm{N/m}}} \right)\left( {0.0440\;{\rm{m}}} \right)\\[0.3cm] &= 241.884816754\;{\rm{N}}\\[0.3cm] &\approx 241.9\;{\rm{N}} \end{align*} {/eq}

Thus, the force needed to compress the spring by {eq}0.0440\;{\rm{m}} {/eq} is {eq}\boxed{\color{blue}{241.9\;{\rm{N}}}}. {/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.