Problem 1.60 — Braking force — constant deceleration from friction

Problem Statement

Solve the Newton’s Laws / mechanics problem: A car of mass $m = 1500\,\text{kg}$ decelerates from $v_0 = 100\,\text{km/h}$ to rest under constant friction force $F$. Stopping distance $s = 50\,\text{m}$. Find $F$ and the braking time $t$. Work-energy theorem: $$F\cdot s = \frac12 mv_0^2$$ $$F = \frac{mv_0^2}{2s} = \frac{1500\times(100/3.6)^2}{

Given Information

• $m = 1500\,\text{kg}$
• $v_0 = 100\,\text{km/h}$
• $s = 50\,\text{m}$

Physical Concepts & Formulas

Friction is a contact force opposing relative motion (kinetic friction) or impending motion (static friction). On an inclined plane, the weight component along the slope is $mg\sin\theta$ and the normal force is $N = mg\cos\theta$, giving maximum static friction $f_{s,\max} = \mu_s mg\cos\theta$. The condition for sliding is $\tan\theta > \mu_s$.

• $f = \mu N$ — kinetic friction force
• $N = mg\cos\theta$ — normal force on incline
• $mg\sin\theta – \mu mg\cos\theta = ma$ — Newton’s 2nd law along incline
• $\tan\theta_c = \mu_s$ — critical angle for sliding

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

$$F\cdot s = \frac12 mv_0^2$$

$$F = \frac{mv_0^2}{2s} = \frac{1500\times(100/3.6)^2}{

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:

$$

$$

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: $m_1 = 3\,\text{kg}$, $m_2 = 5\,\text{kg}$:

$$

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.