Newton's Laws (Calculus-Based)

Newton's Second Law (Differential Form)

$F_{\text{net}} = m\frac{dv}{dt} = m\frac{d^2x}{dt^2}$

Velocity-Dependent Forces (Drag)

Linear drag: $F_d = -bv$

Equation of motion for an object falling with linear drag: $m\frac{dv}{dt} = mg - bv$

Terminal velocity: when $a = 0$: $v_T = mg/b$.

Solution: $v(t) = v_T(1 - e^{-bt/m})$

Position-Dependent Forces (Springs)

Hooke's law: $F = -kx$ $m\frac{d^2x}{dt^2} = -kx \quad \Rightarrow \quad x(t) = A\cos(\omega t + \phi)$ where $\omega = \sqrt{k/m}$. (SHM — covered in detail in the Oscillations topic.)

Solving Motion Problems with Calculus

1. Draw an FBD and identify all forces.
2. Write $\Sigma F = ma$ (or $\Sigma F = m\,dv/dt$).
3. If force depends on $v$: separate variables ($dv$, $dt$) and integrate.
4. If force depends on $x$: use $F = mv\,dv/dx$ or energy methods.

Friction (Review)

• Static: $f_s \leq \mu_s N$
• Kinetic: $f_k = \mu_k N$

On inclines and in complex systems, set up component equations and solve.

Problem: A $2\ \text{kg}$ object falls from rest with drag force $F_d = -4v$. Find $v(t)$ and the terminal velocity.

Solution:

$m\frac{dv}{dt} = mg - bv$ $2\frac{dv}{dt} = 2(9.8) - 4v = 19.6 - 4v$

Terminal velocity: $v_T = mg/b = 2(9.8)/4 = 4.9\ \text{m/s}$.

Separating variables: $\frac{dv}{19.6 - 4v} = \frac{dt}{2}$ $-\frac{1}{4}\ln|19.6 - 4v| = \frac{t}{2} + C$

With $v(0) = 0$: $C = -\frac{1}{4}\ln 19.6$.

$v(t) = 4.9(1 - e^{-2t})$

• Newton's second law becomes a differential equation when forces are variable.
• Drag $F = -bv$: leads to exponential approach to terminal velocity $v_T = mg/b$.
• Position-dependent forces (springs): lead to simple harmonic motion.
• Technique: identify force dependence, separate variables, integrate with initial conditions.