Solution to Question 28
1. Initial Equilibrium State
Initially, both springs are compressed under the weight of the blocks.
- Compression in $k_2$: It supports both masses. $$x_2 = \frac{(m_1 + m_2)g}{k_2}$$
- Compression in $k_1$: It supports mass $m_1$. $$x_1 = \frac{m_1 g}{k_1}$$
2. Final State
The system is lifted until the lower spring $k_2$ is relaxed (force is zero). At this point, Block B is effectively hanging from Block A via spring $k_1$.
- Since $k_2$ provides no support, $k_1$ must support the weight of $m_2$.
- Spring $k_1$ is now extended by $x_1’$. $$x_1′ = \frac{m_2 g}{k_1}$$
3. Calculate Height Increments
Rise of Block B ($\Delta h_2$): B moves from compressed position to the natural length position of $k_2$.
$$\Delta h_2 = x_2 = \frac{(m_1 + m_2)g}{k_2}$$Rise of Block A ($\Delta h_1$): A moves up by the amount B moves, plus the change in length of spring $k_1$ (from compressed $x_1$ to extended $x_1’$).
$$\Delta h_1 = \Delta h_2 + (x_1 + x_1′) = \frac{(m_1 + m_2)g}{k_2} + \left(\frac{m_1 g}{k_1} + \frac{m_2 g}{k_1}\right) = \frac{(m_1 + m_2)g}{k_2} + \frac{(m_1 + m_2)g}{k_1}$$