Chapter 1.10, Problem 65E

To determine

To find the amount of force required to stretch the spring 0.16 meter. Given that, a force of 220 newtons stretches a spring 0.12 meter.

Answer to Problem 65E

  $\text{293}\text{newtons}$

Explanation of Solution

Given:

A force of 220 newtons stretches a spring 0.12 meters.

Concept used:

Hooke’s law: It states that the distance of a spring stretches from its natural, or equilibrium, length varies directly as the applied force on the spring.

Calculation:

According to Hooke’s law, distance of a spring stretched from its natural position is directly

proportional to applied force on the spring.

  $\begin{array}{l}\Rightarrow F\propto d\\ \Rightarrow F=kd\mathrm{...}(i)[\text{Removing proportionality}]\end{array}$

Where k is proportionality constant

Also, a force of 220 newtons stretches a spring 0.12 meters.

  $\begin{array}{l}\Rightarrow F=kd\\ \Rightarrow 220=k(0.12)\\ \Rightarrow \frac{220}{0.12}=k\\ \Rightarrow \text{1833}\text{.33}=k\\ \Rightarrow k=\text{1833}\text{.33}\mathrm{...}(ii)\end{array}$

Substituting equation (ii) in equation (i),

  $\Rightarrow F=\text{1833}\text{.33}d$

If the distance is 0.16 meter, the force required is

  $\begin{array}{l}\Rightarrow F=\text{1833}\text{.33}d\\ \Rightarrow F=\text{1833}\text{.33}(0.16)\\ \Rightarrow F=\text{293}\text{.3328}\\ \Rightarrow F\approx \text{293}\end{array}$

Conclusion:

Therefore, the amount of force required to stretch the spring 0.16 meter is $\text{293}\text{newtons}.$