Solution to Question 14
Problem Analysis:
We have a system with two blocks: Block A (mass $m$) and Block B (mass $2m$). The string is pulled with a constant force $F$. We need to find the speeds of blocks A and B when the free end of the thread acquires a speed $v$.
1. Constraint Relationship (Virtual Work Method)
Since the thread is inextensible, the total work done by the tension forces on the system is zero ($\sum \vec{T} \cdot \vec{v} = 0$).
- At the hand: The tension $T$ acts upwards (along the string), and the displacement is upwards. Rate of work = $+T v$.
- At Block B: The pulley holding B has two string segments pulling it up. The total upward force is $2T$. Block B moves with velocity $v_B$ upwards. Rate of work = $-2T v_B$ (tension pulls up, we consider work done by string on the system essentially accounts for string shortening).
- At Block A: The string pulls Block A upwards with tension $T$. Block A moves with velocity $v_A$ upwards. Rate of work = $-T v_A$.
Summing these up: $$T v – 2T v_B – T v_A = 0$$ Dividing by $T$: $$v = 2v_B + v_A \quad \dots(1)$$
2. Dynamics and Acceleration
Let’s analyze the forces to find the relationship between $a_A$ and $a_B$.
- For Block A (mass $m$): $$T – mg = m a_A \implies a_A = \frac{T}{m} – g$$
- For Block B (mass $2m$): The upward force is $2T$. $$2T – 2mg = (2m) a_B \implies a_B = \frac{2T – 2mg}{2m} = \frac{T}{m} – g$$
Comparing the two expressions, we see that: $$a_A = a_B$$ Since both blocks start from rest and have the same acceleration, their velocities will always be equal: $$v_A = v_B$$
3. Final Calculation
Substitute $v_A = v_B$ into the constraint equation (1): $$v = 2v_B + v_B$$ $$v = 3v_B$$ $$v_B = \frac{v}{3}$$ And since $v_A = v_B$: $$v_A = \frac{v}{3}$$