Problem Statement
Solve the work-energy problem: A charge of $2\mu$C is moved from a point where potential is 100 V to a point where potential is 20 V. Find the work done by the external agent and by the electric force. Work by external agent: $W_{ext} = q(V_B – V_A)$ Work by electric force: $W_E = q(V_A – V_B)$ Step 1: $q = 2\times10^{-6}$ C, $V_
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
Full substitution shown in the steps above.
Answer
$$\boxed{v_f = \sqrt{2g h}}$$
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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