Consider the spring of relaxed length $l$ placed on a frictionless horizontal floor. A force $F$ is applied at one end. The tension $T(x)$ varies linearly from $0$ at the free end to $F$ at the pulled end: $$T(x) = F \frac{x}{l}$$

The glue breaks only where the tension exceeds the breaking strength $F_b$.

• If $T(x) < F_b$: The turns are held together by glue. Since they "almost touch," this portion acts as a rigid body with zero extension.
• If $T(x) > F_b$: The glue breaks, and this portion extends elastically.

The critical position $x_c$ where the glue breaks is: $$ F \frac{x_c}{l} = F_b \implies x_c = \frac{F_b}{F} l $$

We integrate the extension over the region $x_c$ to $l$. The extension $d\delta$ of a small element $dx$ is $d\delta = \frac{T(x) dx}{kl}$. $$ \Delta l = \int_{x_c}^{l} \frac{F(x/l)}{kl} dx = \frac{F}{kl^2} \int_{x_c}^{l} x \, dx $$ $$ \Delta l = \frac{F}{2kl^2} [l^2 – x_c^2] = \frac{F}{2k} \left[ 1 – \left(\frac{F_b}{F}\right)^2 \right] = \frac{F^2 – F_b^2}{2kF} $$

As established, the spring only extends in the region where Tension $> F_b = 100$ N. The tension varies linearly from $0$ to $F=200$ N over the length $l=1$ m. The “active” portion of the spring is the segment where tension rises from $100$ N to $200$ N.

Since $100$ N is exactly half of $200$ N, this active segment corresponds to the outer half of the spring.
Length of active segment: $l_{active} = \frac{l}{2} = 0.5 \text{ m}$.

The spring constant is inversely proportional to length ($k \propto 1/L$). Since the active segment is half the original length ($l/2$), its stiffness $k’$ is double the original stiffness $k$. $$ k’ = 2k = 2(500) = 1000 \text{ N/m} $$

Since the tension varies linearly with position ($T \propto x$), we can calculate the extension of the active segment using the Average Tension acting on it.

Force at start of active segment: $F_{start} = F_b = 100 \text{ N}$
Force at end of active segment: $F_{end} = F = 200 \text{ N}$

$$ \langle F \rangle = \frac{F_{start} + F_{end}}{2} = \frac{100 + 200}{2} = 150 \text{ N} $$

Now, apply Hooke’s Law for the active segment: $$ \Delta l = \frac{\langle F \rangle}{k’} $$ $$ \Delta l = \frac{150}{1000} = 0.15 \text{ m} $$

$$ L_{final} = l_{original} + \Delta l = 1.00 + 0.15 = 1.15 \text{ m} $$