# A spring is hung from the ceiling then a mass of 10 kg is attached to its other end. This causes...

## Question:

A spring is hung from the ceiling then a mass of 10 kg is attached to its other end. This causes the spring to stretch 3 cm.

a) What is the spring constant of the spring?

b) What is the "potential energy" stored in the spring?

## Spring Force:

- The restoring force developed in the spring is equal to the external force applied to it. This restoring force is directly proportional to the amount of stretch-induced in the spring. The constant of proportionality is known as the
*spring constant.*

The force in a spring is given as:

$$\hspace{1cm}F= kx$$

The potential energy stored in the spring is given as:

$$\hspace{1cm}U= \frac{1}{2}kx^2$$

In this formula,

- $$F$$ is the restoring force.

- $$U$$ is the potential energy stored in the spring.

- $$k$$ is the spring constant and

- $$x$$ is the change in the length of the spring about its natural length.

## Answer and Explanation: 1

**We have the following given data**

$$\begin{align} \\ m &=10~\rm{kg } \\[0.3cm] x &=3 ~\rm{cm } \\[0.3cm] k &= \, ? && \left[ \ \text{ Spring constant} \ \right] \\[0.3cm] U &= \, ? && \left[ \ \text{ Potential energy stored in the spring} \ \right] \\[0.3cm] \end{align}\\ $$

**Solution**

#### Part a.

The restoring force developed in the spring is equal to the external force applied to it. This restoring force is directly proportional to the amount of stretch-induced in the spring. The constant of proportionality is known as the *spring constant. *

The force in a spring is given as:

$$\hspace{1cm}F= kx$$

In this formula,

- $$F$$ is the restoring force

- $$k$$ is the spring constant and

- $$x$$ is the change in the length of the spring about its natural length.

Plugging in our values, we have

$$\begin{align} \\ ~\text{Restoring force in the spring } ~ &= ~\text{Weight of the hanging mass } ~\\[0.3cm] F_{Sping}&=mg \\[0.3cm] kx&=mg \\[0.3cm] k&=\frac{mg}{x} && \left[ \ \text{ Extracting }~ k \ \right] \\[0.3cm] k&=\frac{(10)(9.81)}{0.03} \\[0.3cm] k&=3270 ~\rm{N/m } \\[0.3cm] \end{align}\\ $$

#### Part b.

The potential energy stored in the spring is given as:

$$\hspace{1cm}U= \frac{1}{2}kx^2$$

In this formula,

- $$U$$ is the potential energy stored in a spring

- $$k$$ is the spring constant and

- $$x$$ is the change in the length of the spring about its natural length.

Plugging in our values, we have

$$\begin{align} \\ U&= \frac{1}{2}kx^2 \\[0.3cm] U&= \frac{1}{2}(3270 ) (0.03)^2 \\[0.3cm] U&=1.4715~\rm{J} \\[0.3cm] U&\approx 1.47~\rm{J} && \left[ \ \text{Correct to two decimal places } \ \right] \\[0.3cm] \end{align}\\ $$

**Therefore, the spring constant and the potential energy stored in the spring is **{eq}\displaystyle \boxed{\color{blue} { \boldsymbol{ \matrix{\begin{align} \\
k&=3270 ~\rm{N/m } \\[0.3cm]
U&\approx 1.47~\rm{J}
\end{align}\\ } } }}
{/eq}

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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.