Problem Statement
Solve the Newton’s Laws / mechanics problem: A 2 kg block moving at 6 m/s on a frictionless surface hits a spring ($k = 400$ N/m). Find the maximum compression of the spring. Energy conservation: $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$ Step 1: $x = v\sqrt{m/k} = 6\sqrt{2/400} = 6\sqrt{0.005} = 6 \times 0.0707 = 0.424$ m. $$\boxed{x_{max} = v\sqrt{
Given Information
- Mass(es), forces, angles, and coefficients of friction as given
- $g = 9.8\,\text{m/s}^2$ (acceleration due to gravity)
Physical Concepts & Formulas
Newton’s three laws form the complete foundation of classical mechanics. The second law $\vec{F}_{\text{net}} = m\vec{a}$ is the workhorse: draw a free-body diagram for each object, resolve forces into components along chosen axes, and write $\sum F_x = ma_x$, $\sum F_y = ma_y$. For systems connected by strings over pulleys, the tension is common to both sides of a massless string. Friction force $f = \mu N$ opposes relative sliding and is proportional to the normal force, not the contact area.
- $\vec{F}_{\text{net}} = m\vec{a}$ — Newton’s second law
- $f_k = \mu_k N$ — kinetic friction
- $f_{s,\max} = \mu_s N$ — maximum static friction
- $N = mg\cos\theta$ — normal force on incline of angle $\theta$
- $a = g\sin\theta – \mu_k g\cos\theta$ — acceleration on rough incline
Step-by-Step Solution
Step 1 — Free-body diagram: Draw all forces on each object separately.
Step 2 — Choose coordinate axes: Align one axis along the direction of motion.
Step 3 — Apply Newton’s 2nd Law component by component:
$$\sum F_x = ma_x\quad,\quad \sum F_y = ma_y = 0\quad\text{(if no vertical acceleration)}$$
Step 4 — Constraint equations: For Atwood machine: $a_1 = -a_2 = a$, same tension $T$.
Step 5 — Solve the system of equations for $a$ and $T$.
Worked Calculation
Substituting all values with units:
Atwood machine: $m_1 = 3\,\text{kg}$, $m_2 = 5\,\text{kg}$:
$$a = \frac{(m_2-m_1)g}{m_1+m_2} = \frac{(5-3)\times9.8}{8} = \frac{19.6}{8} = 2.45\,\text{m/s}^2$$
$$T = \frac{2m_1 m_2 g}{m_1+m_2} = \frac{2\times3\times5\times9.8}{8} = \frac{294}{8} = 36.75\,\text{N}$$
Answer
$$\boxed{a = \dfrac{(m_2-m_1)g}{m_1+m_2}}$$
Physical Interpretation
In the Atwood machine, if $m_1 = m_2$ then $a = 0$ (equilibrium). The acceleration grows as the mass difference increases and approaches $g$ when one mass is negligible. The tension is always between $m_1 g$ and $m_2 g$ — the string must support the lighter side while slowing the heavier one. Real pulleys have mass (moment of inertia) which reduces $a$ slightly.
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