Animated Solution for Physics - Work, Energy, and Power: A spring of spring constant $5\times 10^3$ N/m is stretched initially by $5$ cm from the unstretched position. Then, the work required to stretch it further by another $5$ cm is

Imagine you are pulling a heavy resistance band. At first, it stretches easily, but the further you pull, the harder it fights back. This is the essence of Hooke's Law, which states that the restoring force of a spring is directly proportional to its extension: $F=kx$.

Because the force is not constant, we cannot simply use the basic work formula $W=F\times d$. Instead, we must calculate the work done as the area under the Force-Extension ($F-x$) graph. For a spring starting from its natural length, this area forms a triangle, giving us the famous elastic potential energy formula:

The problem states that the spring is initially stretched by $5\text{ cm}$. Before we plug anything into our formula, we must ensure our units are in standard SI format. So, $x_1=5\text{ cm}=5\times 10^{-2}\text{ m}$.

Let's calculate the work done for this first stretch, which we will call $W_1$:

Why We Can't Just Use $x=5\text{ cm}$ Again

Here is where many students make a critical error. The question asks for the work required to stretch the spring further by another $5\text{ cm}$. It is incredibly tempting to just say, "Well, it's another $5\text{ cm}$ stretch, so the work must be another $6.25\text{ J}$."

When you start the second stretch, the spring is already stretched. You are no longer starting from a force of zero; you are starting against a significant restoring force. Geometrically, the work done for this second phase is not a small triangle starting from the origin, but rather a large trapezium sitting further down the $x$-axis.

To find the work for the second phase, the safest method is to calculate the total work done from the very beginning to the final position, and then subtract the work we already did.

The final extension from the natural length is $x_2=5\text{ cm}+5\text{ cm}=10\text{ cm}=10\times 10^{-2}\text{ m}$.

Let's calculate the total work, $W_2$:

Now, to find the additional work ($\Delta W$) required just for that second $5\text{ cm}$ stretch, we subtract the initial work from the total work:

Quadratic Scaling

Notice the profound physical reality hidden in these numbers. The first $5\text{ cm}$ stretch required $6.25\text{ J}$ of energy. The next $5\text{ cm}$ stretch required $18.75\text{ J}$ of energy.

Even though the distance was exactly the same, the second stretch required three times as much work! This is the hallmark of a quadratic relationship ($W\propto x^2$). As you stretch a spring further, the energy required doesn't just increase; it accelerates.