Block Floating 08011F

1. **Problem statement:** A rectangular block with density $750\ \text{kg m}^{-3}$ floats in water with density $1000\ \text{kg m}^{-3}$. We need to find: a) The fraction of the block's vertical height submerged in water. b) The initial acceleration of the block when fully submerged and released. 2. **Relevant formulas and principles:** - Archimedes' principle states that the buoyant force equals the weight of the displaced fluid. - For a floating object, the weight of the object equals the buoyant force. - Density $\rho = \frac{\text{mass}}{\text{volume}}$. - Weight $W = mg = \rho V g$ where $V$ is volume and $g$ is acceleration due to gravity. 3. **Part (a): Fraction submerged** Let the total height of the block be $H$ and the submerged height be $h$. Volume submerged $V_{sub} = A h$ where $A$ is the cross-sectional area. Total volume $V = A H$. Since the block floats, weight equals buoyant force: $$\rho_{block} V g = \rho_{water} V_{sub} g$$ Cancel $g$ and $A$: $$\rho_{block} H = \rho_{water} h$$ Solve for fraction submerged $\frac{h}{H}$: $$\frac{h}{H} = \frac{\rho_{block}}{\rho_{water}} = \frac{750}{1000} = 0.75$$ 4. **Part (b): Initial acceleration when fully submerged** When fully submerged, the buoyant force is: $$F_b = \rho_{water} V g$$ Weight of block: $$W = \rho_{block} V g$$ Net force upwards: $$F_{net} = F_b - W = (\rho_{water} - \rho_{block}) V g$$ Using Newton's second law: $$F_{net} = m a = \rho_{block} V a$$ So, $$\rho_{block} V a = (\rho_{water} - \rho_{block}) V g$$ Cancel $V$: $$\rho_{block} a = (\rho_{water} - \rho_{block}) g$$ Solve for $a$: $$a = \frac{(\rho_{water} - \rho_{block}) g}{\rho_{block}} = \frac{(1000 - 750) g}{750} = \frac{250 g}{750} = \frac{1}{3} g$$ Using $g = 9.8\ \text{m/s}^2$: $$a = \frac{1}{3} \times 9.8 = 3.27\ \text{m/s}^2$$ upwards. **Final answers:** - a) Fraction submerged = $0.75$ - b) Initial acceleration upwards = $3.27\ \text{m/s}^2$