Physics Solution: Work Done on a Rope
The work done by an external agent is equal to the change in the potential energy of the system: $W_{ext} = \Delta U$.
Step-by-Step Derivation
1. Define Potential Energy States
Let the reference level for gravitational potential energy ($y=0$) be the horizontal line connecting the nails.
- State C (Equilibrium): The rope hangs in a curve. Let its potential energy be $U_{eq}$. Note that $U_{eq}$ is negative because the mass hangs below the reference line.
- State D (V-shape): The midpoint is pulled down. The COM is at depth $Y$. $U_D = -MgY$.
- State E (Inverted V-shape): The midpoint is pulled up. The COM is at height $Y$. $U_E = +MgY$.
2. Use the Given Work $W$
The total work done for the two pulling processes is given as $W$.
$$W = (U_D – U_{eq}) + (U_E – U_{eq})$$
Substituting the symmetry condition ($U_D + U_E = 0$):
$$W = (U_D + U_E) – 2U_{eq}$$
$$W = 0 – 2U_{eq} \implies U_{eq} = -\frac{W}{2}$$
3. Analyze the Final State F
In state F, the midpoint is lifted to the level of the nails. The rope forms two identical smaller loops side-by-side.
- Scaling: Each small loop has half the span and half the total length of the original rope.
- Geometric Similarity: For a hanging rope under gravity, if you scale the length and span by half, the vertical sag also scales by half.
- Potential Energy Scaling: $$U \propto \text{Mass} \times \text{Depth of COM}$$ For one small loop: Mass is $M/2$. Depth is $y_{com}/2$. $$U_{small} \propto \left(\frac{M}{2}\right) \left(\frac{y_{com}}{2}\right) = \frac{1}{4} (M y_{com})$$ So, $U_{small} = \frac{1}{4} U_{eq}$.
- Since there are two such loops: $U_F = 2 \times U_{small} = \frac{1}{2} U_{eq}$.
4. Calculate Required Work
The work done to move from equilibrium (C) to state F is:
$$\text{Work} = U_F – U_{eq}$$
$$\text{Work} = \frac{1}{2} U_{eq} – U_{eq} = -\frac{1}{2} U_{eq}$$
Substituting $U_{eq} = -W/2$:
$$\text{Work} = -\frac{1}{2} \left(-\frac{W}{2}\right) = \frac{W}{4}$$