Animated Solution for Physics - Work, Energy, and Power: A spring of force constant $800\text{ N/m}$ has an extension of $5\text{ cm}$. The work done in extending it from $5\text{ cm}$ to $15\text{ cm}$ is

Imagine you are holding a heavy-duty spring. When you first start pulling it, it feels relatively easy to stretch. But as you pull it further, it fights back harder and harder. This intuitive feeling is perfectly captured by Hooke's Law, which states that the restoring force exerted by a spring is directly proportional to its extension: $F=kx$.

Because the force is constantly changing as you stretch the spring, you cannot simply use the basic work formula $W=F\times d$. Instead, we must calculate the work done by finding the area under the Force vs. Displacement graph, or by using calculus to integrate the force over the distance.

The work done by an external force in stretching a spring from an initial position $x_1$ to a final position $x_2$ is stored as elastic potential energy in the spring. Mathematically, this is expressed as:

Evaluating this integral gives us our master equation for the work done:

Before we plug numbers into our equation, we must ensure all our units are in the standard SI system. This is a classic trap where many students lose marks! The spring constant is given as $k=800\text{ N/m}$.

The initial extension is $5\text{ cm}$, which we must convert to meters:

The final extension is $15\text{ cm}$, which becomes:

Now, let's substitute these values into our master equation:

Squaring decimals can sometimes lead to silly arithmetic errors. Instead, we can use a clever algebraic identity: $a^2-b^2=(a-b)(a+b)$. Applying this to our equation makes the mental math a breeze:

Multiplying the terms inside the bracket gives $0.02$. Finally, we multiply this by $400$:

The total work done in extending the spring from $5\text{ cm}$ to $15\text{ cm}$ is exactly $8\text{ Joules}$.