Question

Consider the following system of massless and frictionless pulleys, ropes, and springs:

Initially, a block of mass m is attached to the end of a rope, and the system is in equilibrium. Next the block is doubled in mass, and the system is allowed to come to equilibrium again. During the transition between these equilibria, how far does the end of the rope (where the block is suspended) move ?

• A. $\frac{11 \text{mg}}{12 \text{k}}$
• B. $\frac{13 \text{mg}}{12 \text{k}}$
• C. $\frac{7 \text{mg}}{6 \text{k}}$
• D. $\frac{11 \text{mg}}{6 \text{k}}$

Answer 1

The correct answer is $\frac{8mg}{3k}$, which is not among the provided options. The options might be flawed or there may be a non-standard interpretation of the system.

Answer 2

To solve this problem, we analyze the forces and displacements in the pulley system.

1. Analyze forces and displacements in the system:

   Let $T$ be the tension in the rope. Since the pulleys and ropes are massless and frictionless, the tension is uniform throughout the rope.

   • Block of mass M: The block is in equilibrium, so the upward tension balances its weight: $T = Mg$

   • Upper Pulley (P1): This pulley is connected to the ceiling by a spring with constant $3k$. The rope passes over P1. The two segments of the rope pull P1 downwards. The spring $3k$ pulls P1 upwards. Let $x_1$ be the downward displacement of P1 from its equilibrium position. For equilibrium of P1: $2T = 3k x_1 \implies x_1 = \frac{2T}{3k}$

   • Lower Pulley (P2): This pulley is connected to the ground by a spring with constant $k$. The rope passes under P2. The two segments of the rope pull P2 upwards. For the system to be in equilibrium with the mass pulling down, the spring $k$ must exert a downward force on P2. This implies the spring is either compressed and pushing P2 down, or it's stretched and attached to a point above P2 (which is not how it's drawn, but is the only way for the system to work in equilibrium). Let $x_2$ be the downward displacement of P2 from its equilibrium position. For equilibrium of P2: $2T = k x_2 \implies x_2 = \frac{2T}{k}$

2. Relate the displacement of the mass to the displacements of the pulleys:

   Let $y_0$ be the fixed position of the rope at the ceiling. Let $y_1$ be the position of P1, $y_2$ be the position of P2, and $y_m$ be the position of the mass, all measured downwards from their initial (unstretched spring) positions.

   The total length of the rope, $L$, is constant. The rope segments are:

   • From the fixed point ($y_0$) to P1: $y_1 - y_0$

   • From P1 to P2: $y_1 - y_2$ (assuming P1 is above P2, which is true)

   • From P2 to the mass: $y_m - y_2$

   The total length of the rope is L = (y_1 - y_0) + (y_1 - y_2) + (y_m - y_2) + \text{constant_for_pulley_radii} $L = 2y_1 - 2y_2 + y_m + \text{constant}$.

   Since $L$ is constant, the change in length $\Delta L = 0$. $0 = 2 \Delta y_1 - 2 \Delta y_2 + \Delta y_m$.

   Let $x_m = \Delta y_m$, $x_1 = \Delta y_1$, $x_2 = \Delta y_2$ be the downward displacements. So, $x_m = 2x_2 - 2x_1$.

3. Substitute forces into the displacement relation:

   Substitute the expressions for $x_1$ and $x_2$ in terms of $T$: $x_m = 2 \left(\frac{2T}{k}\right) - 2 \left(\frac{2T}{3k}\right)$ $x_m = \frac{4T}{k} - \frac{4T}{3k}$ $x_m = 4T \left(\frac{1}{k} - \frac{1}{3k}\right)$ $x_m = 4T \left(\frac{3-1}{3k}\right)$ $x_m = 4T \left(\frac{2}{3k}\right)$ $x_m = \frac{8T}{3k}$

   Since $T = Mg$, the total displacement of the mass from its reference position is: $x_m = \frac{8Mg}{3k}$

4. Calculate the displacement during the transition:

   • Initial Equilibrium (Mass = m): The tension is $T_1 = mg$. The initial displacement of the block is $x_{m1} = \frac{8 mg}{3k}$.

   • Final Equilibrium (Mass = 2m): The tension is $T_2 = 2mg$. The final displacement of the block is $x_{m2} = \frac{8 (2mg)}{3k} = \frac{16 mg}{3k}$.

   • Distance moved during transition: The distance moved by the end of the rope is the difference between the final and initial displacements: $\Delta x = x_{m2} - x_{m1}$ $\Delta x = \frac{16 mg}{3k} - \frac{8 mg}{3k}$ $\Delta x = \frac{8 mg}{3k}$

Conclusion:

The calculated displacement is $\frac{8mg}{3k}$. This value is not present in any of the given options. This suggests a potential error in the question's options or a very specific, non-standard interpretation of the pulley-spring system's diagram. However, based on standard physics principles for such systems, this derivation is consistent.

Therefore, the final answer is $\frac{8mg}{3k}$.