Problem Statement
Solve the fluid mechanics problem: Find the work done to blow a soap bubble from radius $R_1=0$ to $R_2 = 3.0\ \text{cm}$. ($\sigma=0.040\ \text{N/m}$) A soap bubble has two surfaces. The surface energy is $E = \sigma\times2\times4\pi R^2 = 8\pi\sigma R^2$. Work done to create the bubble (against surface tension, ignoring gas compres
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Newton’s second law $\mathbf{F}_\text{net} = m\mathbf{a}$ is the fundamental relation between net force and acceleration. For systems of connected objects (Atwood machine, blocks on inclines), each body is treated separately with a free-body diagram, and the constraint equations (same rope length, etc.) link the accelerations.
- $\mathbf{F}_{\text{net}} = m\mathbf{a}$ — Newton’s second law
- Atwood: $a = (m_1-m_2)g/(m_1+m_2)$, $T = 2m_1m_2g/(m_1+m_2)$
- $f_k = \mu_k N$ — kinetic friction
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$$
$$v = \sqrt{2gh} = \sqrt{2\times9.8\times2} = \sqrt{39.2} \approx 6.26\,\text{m/s}$$
$$\boxed{v_{\text{efflux}} = \sqrt{2gh}}$$
Answer
$$\boxed{v_{\text{efflux}} = \sqrt{2gh}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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