Animated Solution for Physics - Properties of Solids and Liquids: A wire suspended vertically from one of its ends is stretched by attaching a weight of $200\text{ N}$ to the lower end. The weight stretches the wire by $1\text{ mm}$. Then, the elastic energy stored in the wire is

Imagine a wire hanging freely from a rigid support on the ceiling. We attach a heavy weight of $200\text{ N}$ to its lower end. This weight acts as a downward force, pulling the wire and causing it to stretch by exactly $1\text{ mm}$.

Our goal is to find the elastic potential energy stored within this stretched wire. When a material is deformed, the work done against the internal restoring forces is stored as potential energy, much like stretching a rubber band.

To find this energy, we start with the fundamental formula for the elastic potential energy stored in a deformed body:

At first glance, this might look intimidating because we aren't given the cross-sectional area or the original length of the wire. But let's break this formula down into its basic components and see what happens.

We know that: - Stress is the restoring force per unit area: $\text{Stress}=\frac{F}{A}$ - Strain is the fractional change in length: $\text{Strain}=\frac{\Delta l}{L}$ - Volume of the cylindrical wire is its area times its length: $\text{Volume}=A\cdot L$

Let's substitute these definitions back into our master equation:

Notice how beautifully the math works out! The cross-sectional area $A$ in the denominator of stress cancels out with the $A$ in the volume. Similarly, the original length $L$ in the denominator of strain cancels out with the $L$ in the volume.

We are left with a remarkably simple and elegant formula:

Now, we just need to plug in our given values. The force applied is $F=200\text{ N}$. The extension is $\Delta l=1\text{ mm}$.

Crucial Step: We must convert the extension into standard SI units (meters) to ensure our final energy is in Joules. So, $\Delta l=10^{-3}\text{ m}$.

Substituting these into our simplified formula:

The elastic energy stored in the wire is exactly $0.1\text{ J}$.

Why the Half?

You might be wondering, why is there a factor of $\frac{1}{2}$ in the formula? Why isn't the work done simply Force $\times$ Distance ($F\times \Delta l$)?

This is a profound physical insight. The restoring force in the wire is not constant. When you first start stretching the wire, the restoring force is zero. As the wire stretches further, the force increases linearly until it reaches the maximum value of $F$.

Because the force increases linearly, the work done is the area under the Force-Extension graph, which forms a triangle. The area of a triangle is $\frac{1}{2}\times \text{base}\times \text{height}$, which gives us our $\frac{1}{2}F\Delta l$. Alternatively, you can think of the $\frac{1}{2}$ as giving us the average force $\left(\frac{0+F}{2}\right)$ exerted during the stretching process.