Problem Statement
Solve the Newton’s Laws / mechanics problem: Two equal charges are placed at a separation of 1.0 m. What is the charge if the Coulomb force between them is 0.1 N? Coulomb’s Law for equal charges: $F = \dfrac{kq^2}{r^2}$ Step 1: Given $F = 0.1$ N, $r = 1.0$ m, $k = 9\times10^9$ N m$^2$/C$^2$. Step 2: Solve for $q$: $$q^2 = \frac{Fr^2}{k} = \fra
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Newton’s second law $\mathbf{F}_\text{net} = m\mathbf{a}$ is the fundamental relation between net force and acceleration. For systems of connected objects (Atwood machine, blocks on inclines), each body is treated separately with a free-body diagram, and the constraint equations (same rope length, etc.) link the accelerations.
- $\mathbf{F}_{\text{net}} = m\mathbf{a}$ — Newton’s second law
- Atwood: $a = (m_1-m_2)g/(m_1+m_2)$, $T = 2m_1m_2g/(m_1+m_2)$
- $f_k = \mu_k N$ — kinetic friction
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$q^2 = \frac{Fr^2}{k} = \fra
Given Information
- Mass(es), forces, angles, and coefficients of friction as given
- $g = 9.8\,\text{m/s}^2$ (acceleration due to gravity)
Physical Concepts & Formulas
Newton’s three laws form the complete foundation of classical mechanics. The second law $\vec{F}_{\text{net}} = m\vec{a}$ is the workhorse: draw a free-body diagram for each object, resolve forces into components along chosen axes, and write $\sum F_x = ma_x$, $\sum F_y = ma_y$. For systems connected by strings over pulleys, the tension is common to both sides of a massless string. Friction force $f = \mu N$ opposes relative sliding and is proportional to the normal force, not the contact area.
- $\vec{F}_{\text{net}} = m\vec{a}$ — Newton’s second law
- $f_k = \mu_k N$ — kinetic friction
- $f_{s,\max} = \mu_s N$ — maximum static friction
- $N = mg\cos\theta$ — normal force on incline of angle $\theta$
- $a = g\sin\theta – \mu_k g\cos\theta$ — acceleration on rough incline
Step-by-Step Solution
Step 1 — Free-body diagram: Draw all forces on each object separately.
Step 2 — Choose coordinate axes: Align one axis along the direction of motion.
Step 3 — Apply Newton’s 2nd Law component by component:
$$
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Step 4 — Constraint equations: For Atwood machine: $a_1 = -a_2 = a$, same tension $T$.
Step 5 — Solve the system of equations for $a$ and $T$.
Worked Calculation
Substituting all values with units:
Atwood machine: $m_1 = 3\,\text{kg}$, $m_2 = 5\,\text{kg}$:
$$
$$
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Answer
$$\boxed{a = \dfrac{(m_2-m_1)g}{m_1+m_2}}$$
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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