# A spring has an unstretched length of 10 cm. It exerts a restoring force F when stretched to a...

## Question:

A spring has an unstretched length of 10 cm. It exerts a restoring force F when stretched to a length of 12 cm.

A. For what total stretched length of the spring is its restoring force 3F?

B. At what compressed length is the restoring force 2F?

## Restoring Force:

When an object connects with a spring end and tries to compress or stretch the spring from its original position then, the spring applies a force due to the spring stiffness. This force tries to maintain the original position of the spring. If the value of spring stiffness is more it means the value of this force would be large.

## Answer and Explanation: 1

Given Data

• The un-stretched length of spring is {eq}{x_1} = 10\,\;{\rm{cm}}\;{\rm{ = }}\;{\rm{0}}{\rm{.10}}\;{\rm{m}} {/eq}.
• The value of restoring force is {eq}F {/eq}.
• The stretched length is {eq}{x_2} = 12\;{\rm{cm}}\;{\rm{ = }}\;{\rm{0}}{\rm{.12}}\;{\rm{m}} {/eq}.
• The other restoring force is {eq}{F_2} = 3F {/eq}.
• The one other restoring force is {eq}{F_3} = 2F {/eq}.

(a)

Let us consider the spring constant is {eq}k {/eq}.

The value of the spring constant has a constant value for a spring.

The formula to calculate the spring constant is given by,

{eq}F = k\left( {{x_2} - {x_1}} \right) {/eq}.

Here, {eq}F {/eq} is the restoring force.

Substitute all the known values in the above formula.

{eq}\begin{align*} F &= k\left( {{x_2} - {x_1}} \right)\\ &= k\left( {0.12 - 0.10} \right)\\ &= 0.02k\\ k &= 50F \end{align*} {/eq}

The formula to calculate the total stretched the length of spring when restoring force is {eq}3F {/eq}.

{eq}{F_2} = k\left( {{x_3} - {x_1}} \right) {/eq}.

Here, {eq}{x_3} {/eq} is the stretched length of spring.

Substitute all the known values in the above formula.

{eq}\begin{align*} {F_2} &= k\left( {{x_3} - {x_1}} \right)\\ 3F &= 50F\left( {{x_3} - 0.10} \right)\\ 0.06 &= {x_3} - 0.10\\ {x_3}&=0.16\;{\rm{m}}\\ &=16\;{\rm{cm}} \end{align*} {/eq}

Thus, the stretched length of spring is {eq}16\;{\rm{cm}} {/eq}.

(b)

The formula to calculate the compressed length of spring when the restoring force is {eq}2F {/eq} is given by,

{eq}{F_3} = k\left( {{x_1} - {x_4}} \right) {/eq}.

Here, {eq}{x_4} {/eq} is the compressed length.

Substitute all the known values in the above formula.

{eq}\begin{align*} {F_3} &= k\left( {{x_1} - {x_4}} \right)\\ 2F &= 50F\left( {0.10 - {x_4}} \right)\\ 0.04 &= 0.10 - {x_4}\\ {x_4} &= 0.06\;{\rm{m}}\\ &=6\;{\rm{cm}} \end{align*} {/eq}

Thus, the value of compressed length is {eq}6\;{\rm{cm}} {/eq}.

#### Learn more about this topic:

Hooke's Law & the Spring Constant: Definition & Equation

from

Chapter 4 / Lesson 19
201K

After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.