Problem Solution: Velocity of Suspended Iron Cube
Physical Analysis: We are analyzing the motion of an iron cube suspended by a spring as the water level around it rises at a constant speed $u$. As the water level rises, the submerged volume of the cube increases, increasing the buoyant force. This reduction in net downward force causes the spring to contract, pulling the block upward. We must determine the speed of the block as a function of time.
Step 1: Establishing the Equilibrium Equation
Let $y$ be the upward displacement of the block from its initial equilibrium position (where it just touches the water). Since the water rises slowly (quasi-static process), the net force on the block remains zero throughout the motion.
At any time $t$, the forces acting on the block are:
- Change in Spring Force: Since the block moves up by $y$, the spring compression/extension changes by $y$, resulting in a downward restoring force change of $ky$. (Or simply, the upward spring force decreases by $ky$).
- Buoyant Force ($F_B$): This acts upwards. $F_B = \text{Volume}_{\text{sub}} \cdot \rho_w \cdot g$.
The water level rises with speed $u$, so the level height is $ut$. The bottom of the block is at height $y$. The submerged depth $x$ is: $$ x = ut – y $$ The submerged volume is $V_{\text{sub}} = l^2(ut – y)$.
The equilibrium condition (change in forces) is: $$ \Delta F_{\text{net}} = F_B – k y = 0 $$ $$ \rho_w g l^2 (ut – y) – ky = 0 $$
Step 2: Solving for Velocity
Rearranging the equation to solve for displacement $y$:
$$ \rho_w g l^2 ut – \rho_w g l^2 y – ky = 0 $$ $$ y (k + \rho_w g l^2) = \rho_w g l^2 u t $$ $$ y(t) = \left( \frac{\rho_w g l^2 u}{k + \rho_w g l^2} \right) t $$To find the velocity of the block, we differentiate $y$ with respect to $t$:
$$ v_{\text{block}} = \frac{dy}{dt} = u \left( \frac{\rho_w g l^2}{k + \rho_w g l^2} \right) $$Step 3: Determining the Time Interval
This velocity is constant but valid only while the block is partially submerged. The motion changes once the block is fully submerged ($x = l$).
The condition for partial submersion is $ut – y \leq l$. Substituting our expression for $y$:
$$ ut – u \left( \frac{\rho_w g l^2}{k + \rho_w g l^2} \right) t \leq l $$ $$ ut \left( 1 – \frac{\rho_w g l^2}{k + \rho_w g l^2} \right) \leq l $$ $$ ut \left( \frac{k}{k + \rho_w g l^2} \right) \leq l $$ $$ t \leq \frac{l}{u} \left( \frac{k + \rho_w g l^2}{k} \right) = \frac{l}{u} \left( 1 + \frac{\rho_w g l^2}{k} \right) $$Once fully submerged ($t$ exceeds this value), the buoyant force becomes constant ($F_B = \rho_w g l^3$). Since $F_B$ is constant, the spring extension becomes constant to balance it, implying the block stops moving ($v=0$).