Newton's Second Law

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Newton's Laws of Motion· 14 min read

1. Core Definition of Newton's Second Law★★☆☆☆⏱ 3 min

Newton's second law is the foundational relationship between net force, mass, and acceleration that underpins almost all of classical mechanics, making up 12-18% of the total AP Physics C: Mechanics exam weight. It is the bridge between kinematics (Unit 1) and dynamic analysis of motion, appearing on both multiple-choice and free-response sections.

📘 Definition

Newton's Second Law

\vec{F}_{net} = \frac{d\vec{p}}{dt}, \quad \vec{F}_{net} = m\vec{a} \text{ (constant mass)}

The net external force acting on a body is equal to the time rate of change of the body's linear momentum. For constant mass systems, this simplifies to net force equals mass times the acceleration of the body's center of mass. Force causes acceleration (change in velocity), not velocity itself.

Example:

A constant net force produces constant acceleration, not constant velocity.

2. Vector Decomposition & Inertial Reference Frames★★☆☆☆⏱ 4 min

The vector nature of Newton's second law lets us decompose $F_{n e t}$ and $a$ into components along any orthogonal coordinate system, giving independent equations for each axis: $F_{n e t, x} = m a_{x}$ , $F_{n e t, y} = m a_{y}$ . Newton's second law only holds in inertial reference frames (non-accelerating frames); the Earth's surface is a valid approximation for all AP problems.

📐 Worked Example

A 10 kg box slides down a 30° inclined plane with coefficient of kinetic friction 0.2. Find the acceleration of the box along the incline.

1
Choose a coordinate system aligned with the incline: $+ x$ down the incline, $+ y$ perpendicular to the incline. This leaves acceleration non-zero only along the $x$ -axis.
2
Decompose the weight $m g$ into components:
3
$mg_x = mg\sin\theta, \quad mg_y = -mg\cos\theta$
4
Normal force $N$ points along $+ y$ , and kinetic friction $f_{k} = μ_{k} N$ points along $- x$ (opposing motion). Apply Newton's second law to the $y$ -axis (acceleration $a_{y} = 0$ ):
5
$F_{net,y} = N - mg\cos\theta = 0 \implies N = mg\cos\theta$
6
Apply Newton's second law to the $x$ -axis and substitute $N$ :
7
$F_{net,x} = mg\sin\theta - \mu_k N = ma \\ a = g\left(\sin\theta - \mu_k \cos\theta\right)$
8
Substitute $g = 9.8 m/s^{2}$ and trigonometric values:
9
$a = 9.8\left(0.5 - 0.2 \cdot \frac{\sqrt{3}}{2}\right) \approx 3.2\ \text{m/s}^2$

Exam tip:

Always align one axis of your coordinate system with the direction of acceleration; this eliminates one non-zero acceleration component and reduces the number of equations you need to solve.

3. Connected Objects: System vs Particle Approach★★★☆☆⏱ 4 min

For multiple connected objects, you can use two valid approaches: (1) treat each object as an individual particle, or (2) treat all connected objects as a single system. For the system approach, only external forces contribute to net force; internal forces (tension, contact forces) cancel out via Newton's third law, so you can ignore them. Use the system approach to quickly find acceleration, and the particle approach to find internal forces like tension.

📐 Worked Example

Two blocks of mass $m_{1} = 2 kg$ and $m_{2} = 3 kg$ are connected by a massless string over a massless, frictionless pulley. $m_{1}$ rests on a frictionless horizontal table, and $m_{2}$ hangs vertically. Find the tension in the string connecting the blocks.

1
Use the system approach first to find the acceleration of the whole system. Only the weight of $m_{2}$ is an external force causing acceleration:
2
$F_{net} = m_2 g = (m_1 + m_2)a \\ a = \frac{m_2 g}{m_1 + m_2} = \frac{3(9.8)}{5} = 5.88\ \text{m/s}^2$
3
Tension is an internal force of the system, so switch to the particle approach applied to $m_{1}$ , where tension is the only horizontal force.
4
Apply Newton's second law to $m_{1}$ to solve for $T$ :
5
$T = m_1 a = \frac{m_1 m_2 g}{m_1 + m_2} = \frac{2 \cdot 3 \cdot 9.8}{5} = 11.76\ \text{N} \approx 11.8\ \text{N}$
6
Verify by applying Newton's second law to $m_{2}$ , which gives the same result:
7
$m_2 g - T = m_2 a \implies T = 3(9.8 - 5.88) = 11.76\ \text{N}$

Exam tip:

If the question asks for an internal force like tension or contact force between two connected objects, always apply Newton's second law to the object on which that force acts directly after finding acceleration from the system approach to avoid sign errors.

4. Newton's Second Law for Variable Acceleration★★★★☆⏱ 3 min

AP Physics C: Mechanics regularly tests Newton's second law for variable forces, which requires calculus. The general form $F_{n e t} = \frac{d p}{d t}$ can be rearranged and integrated to solve for velocity or position when force varies with time, position, or velocity.

📐 Worked Example

A 2 kg object moving along the x-axis is acted on by a time-varying force $F (t) = 4 t - 2 t^{2}$ (in Newtons, for $t \geq 0$ ). At $t = 0$ , the object has an initial velocity of $v_{0} = 3 m/s$ . Find the velocity of the object at $t = 2 s$ .

1
Start from the general form for constant mass:
2
$F(t) = m \frac{dv}{dt}$
3
Rearrange to separate variables and integrate with bounds for velocity and time:
4
$\int_{v_0}^{v(2)} dv = \frac{1}{m} \int_0^2 (4t - 2t^2) dt$
5
Evaluate the integrals: the left side equals $v (2) - v_{0}$ . Simplify the right side:
6
$\frac{1}{2} \left[ 2t^2 - \frac{2}{3}t^3 \right]_0^2 = \frac{1}{2} \left(8 - \frac{16}{3}\right) = \frac{4}{3} \approx 1.33\ \text{m/s}$
7
Solve for final velocity:
8
$v(2) = v_0 + \frac{4}{3} \approx 4.33\ \text{m/s}$

Exam tip:

Remember that the integral of net force over time equals change in momentum (the impulse-momentum theorem), which is just the integrated form of Newton's second law. You can use this shortcut to avoid re-deriving the integral for every variable force problem.

5. Exam-Style Concept Check★★★☆☆⏱ 4 min

✓ Quick check

Test your understanding with these AP-style practice questions:

A 5 kg block is pulled across a horizontal frictionless surface by a 10 N force at an angle of 37° above the horizontal. What is the magnitude of the normal force exerted by the surface on the block? (Use $sin 3 7^{\circ} = 0.6$ , $cos 3 7^{\circ} = 0.8$ , $g = 9.8 m/s^{2}$ )
- 49 N
- 43 N
- 55 N
- 39 N
Reveal answer
43 N —
Sum vertical forces (vertical acceleration = 0): $N + F sin θ - m g = 0 ⟹ N = 49 - 6 = 43$ N. 49 N is the result if you ignore the vertical component of the applied force, a common error.
An Atwood machine has two blocks connected by a massless string over a massless frictionless pulley: block 1 has mass $M$ , block 2 has mass $2 M$ . What is the tension $T$ in the string in terms of $M$ and $g$ ?
- $\frac{1}{3} M g$
- $M g$
- $\frac{4}{3} M g$
- $2 M g$
Reveal answer
$\frac{4}{3}Mg$ —
Adding Newton's second law equations for both blocks gives $a = \frac{g}{3}$ . Substituting back gives $T = M g + M a = \frac{4}{3} M g$ .

6. Common Pitfalls

Wrong move:

Counting internal forces when calculating net force for a system of connected objects, e.g., including tension between two blocks when calculating acceleration of the whole system.

Why:

Students confuse internal and external forces, and often double-count forces when first learning the system approach.

Correct move:

When calculating $F_{n e t}$ for a system, only include forces exerted by objects outside the system; cross out any forces between objects inside your system boundary.

Wrong move:

Decomposing the normal force instead of weight for incline problems, leading to an incorrect expression for normal force.

Why:

Students default to decomposing the "non-weight" force out of habit, even when it leads to extra trigonometry and errors.

Correct move:

Always decompose the weight vector into components parallel and perpendicular to the incline, leaving normal force aligned with your perpendicular axis.

Wrong move:

Forgetting that Newton's second law uses net force, not individual force, when solving for acceleration, e.g., saying acceleration of a falling object with air resistance is $g$ instead of $g - F_{d} / m$ .

Why:

Students rush to use $F = ma$ with the first force they see, instead of summing all forces first.

Correct move:

Always draw a full free-body diagram, sum all force components along each axis before setting the sum equal to $ma$ .

Wrong move:

When integrating a variable force to get velocity, forgetting to add the initial velocity to the integral result.

Why:

Students treat the integral of acceleration as velocity, not change in velocity, skipping the constant of integration.

Correct move:

Always write the definite integral with bounds for initial and final velocity, or explicitly add the initial velocity to the integral of acceleration over time.

Wrong move:

Treating the vertical component of an angled applied force as having no effect on normal force on a horizontal surface.

Why:

Students often only consider the horizontal component of the force when calculating normal force, leading to incorrect values.

Correct move:

Always apply Newton's second law to both axes, even if acceleration is zero along one axis, to get the correct normal force.

7. Quick Reference Cheatsheet

Category	Formula	Notes
General Constant Mass	$F_{n e t} = m a$	Applies to any constant mass body/system in inertial frames; $F_{n e t}$ is vector sum of all external forces
General Momentum Form	$F_{n e t} = \frac{d p}{d t}$	Full definition of Newton's second law, used for variable mass or variable acceleration problems
Decomposed Component Form	$F_{n e t, x} = m a_{x}, F_{n e t, y} = m a_{y}$	Vector decomposition into orthogonal axes simplifies solving 2D problems
System of Connected Objects	$F_{n e t, e x t} = M_{t o t a l} a_{c m}$	Internal forces cancel by Newton's third law; only external forces contribute to acceleration
Integrated Form (Time-Varying Force)	$Δ v = \frac{1}{m} \int_{t_{1}}^{t_{2}} F (t) d t$	Derived directly from Newton's second law, used to find velocity change for time-dependent forces
Incline Normal Force	$N = m g cos θ$	Valid when acceleration is only parallel to the incline and no other forces have perpendicular components
Elevator Tension	$T = m (g + a)$ (upward $a$ ), $T = m (g - a)$ (downward $a$ )	Common real-world application; tension equals weight when acceleration is zero

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Connected objects acceleration
2022 · FRQ
Time-varying force problem
2021 · MCQ
Incline with friction problem

What's Next

Newton's second law is the foundation for all dynamic analysis in AP Physics C: Mechanics, so mastering the techniques here is non-negotiable for all upcoming topics. Next you will apply Newton's second law to circular motion, where acceleration is centripetal and net force points toward the center of the circle. Without a solid understanding of vector decomposition and net force calculation from this topic, solving circular motion and all subsequent problems will be extremely difficult. Later, $F_{n e t} = ma$ is the starting point for work and energy, rotational dynamics, and oscillations—every major unit after this builds on this core relationship. It also underpins the momentum and impulse concepts you will learn later, as the momentum form of Newton's second law is the basis for the impulse-momentum theorem.