Step 1: Determine Breaking Tension ($T_{break}$)

The problem states that if force is applied gradually, the cord breaks at an additional force $F_0$. In a gradual (static) process, acceleration is zero, so forces balance.

Forces: Upward tension $T$, Downward gravity $mg$, Downward applied force $F_0$.

$$T_{break} = mg + F_0$$

Step 2: Dynamic Analysis with Constant Force $F$

Now, a constant force $F$ is applied suddenly to the system initially in equilibrium under gravity alone.

• Initial state: Equilibrium. Stretch $x_i = \frac{mg}{k}$. Velocity $v=0$.
• Final state: Maximum extension. Stretch $x_{max}$. Velocity $v=0$ (momentarily).

We apply the Work-Energy Theorem between these two states:

$$W_{gravity} + W_{force} = \Delta U_{spring}$$

Let $\delta$ be the additional extension beyond the initial equilibrium position ($x_{max} = x_i + \delta$).

$$mg(\delta) + F(\delta) = \frac{1}{2}k(x_i + \delta)^2 – \frac{1}{2}kx_i^2$$

Step 3: Solve for Maximum Extension

Expanding the potential energy term:

$$(mg + F)\delta = \frac{1}{2}k(x_i^2 + 2x_i\delta + \delta^2) – \frac{1}{2}kx_i^2$$ $$(mg + F)\delta = k x_i \delta + \frac{1}{2}k\delta^2$$

Since $k x_i = mg$ (initial equilibrium), the $mg\delta$ terms cancel out:

$$F\delta = \frac{1}{2}k\delta^2$$ $$\delta = \frac{2F}{k}$$

This result is standard: a suddenly applied constant force produces twice the displacement of a gradually applied force.

Step 4: Find Minimum Force $F$

The cord breaks if the maximum tension generated in this dynamic process reaches $T_{break}$.

$$T_{max} = k(x_i + \delta) = k x_i + k\delta = mg + k\left(\frac{2F}{k}\right) = mg + 2F$$

Equating $T_{max}$ to $T_{break}$:

$$mg + 2F = mg + F_0$$ $$2F = F_0$$ $$F = \frac{1}{2}F_0$$