Problem Statement
Solve the kinematics problem: A 5 kg block on a table connected to a 3 kg hanging mass has acceleration 3 m/s². Find the coefficient of kinetic friction between block and table. ($g = 10$ m/s²) $m_2 g – \mu_k m_1 g = (m_1+m_2)a$ Step 1: $\mu_k = \dfrac{m_2 g – (m_1+m_2)a}{m_1 g} = \dfrac{3(10) – 8(3)}{5(10)} = \dfrac{30-24}{50}
Given Information
- Initial velocity $u$ (or $v_0$)
- Acceleration $a$ (constant unless stated otherwise)
- Time $t$ or distance $s$ as given
Physical Concepts & Formulas
Kinematics describes motion without reference to its cause. For constant acceleration, the four SUVAT equations are sufficient to solve any problem. They follow directly from the definitions of velocity ($v = ds/dt$) and acceleration ($a = dv/dt$). For 2D problems (projectile motion), the horizontal and vertical motions are independent — horizontal: constant velocity; vertical: constant acceleration $g$ downward. Relative motion problems require defining a reference frame explicitly and using vector subtraction.
- $v = u + at$
- $s = ut + \tfrac{1}{2}at^2$
- $v^2 = u^2 + 2as$
- $s = \tfrac{1}{2}(u+v)t$
- Range of projectile: $R = \dfrac{u^2\sin 2\theta}{g}$
- Max height: $H = \dfrac{u^2\sin^2\theta}{2g}$
Step-by-Step Solution
Step 1 — List knowns and unknown: $u$, $v$, $a$, $s$, $t$ — identify which three are known.
Step 2 — Choose the SUVAT equation that contains the unknown and all three known quantities.
Step 3 — Substitute and solve algebraically.
Step 4 — For 2D: Decompose $\vec{u}$ into $u_x = u\cos\theta$, $u_y = u\sin\theta$. Solve $x$ and $y$ separately.
Worked Calculation
Substituting all values with units:
Projectile at $u = 20\,\text{m/s}$, $\theta = 30°$:
$$R = \frac{u^2\sin 2\theta}{g} = \frac{400\times\sin 60°}{9.8} = \frac{400\times0.866}{9.8} = \frac{346.4}{9.8} \approx 35.3\,\text{m}$$
$$H = \frac{u^2\sin^2\theta}{2g} = \frac{400\times0.25}{19.6} = \frac{100}{19.6} \approx 5.1\,\text{m}$$
Answer
$$\boxed{R = \dfrac{u^2\sin 2\theta}{g},\quad H = \dfrac{u^2\sin^2\theta}{2g}}$$
Physical Interpretation
Maximum range occurs at $\theta = 45°$ where $\sin 90°=1$. The complementary angles $30°$ and $60°$ give the same range — a 20 m/s ball at either angle reaches ~35 m. Athletes intuitively throw at 45° for distance. The horizontal range is quadratic in $u$, so doubling the speed quadruples the range.
Leave a Reply