Newton's Second Law — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Net force definition, vector form of Newton’s second law, mass vs weight, accelerated motion on inclines, connected object systems, free-body diagram application, and problem-solving for single and multi-object systems.

You should already know: How to draw free-body diagrams for forces acting on a rigid object. How to resolve vectors into perpendicular components. Newton’s first and third laws of motion.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Newton's Second Law?

Newton's Second Law is the core quantitative relationship between force, mass, and acceleration that underpins all of classical dynamics. It is the highest-weight topic in Unit 2 (Dynamics) for AP Physics 1, making up roughly 12-18% of the total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as the foundation for longer FRQ problems that connect to kinematics, energy, or circular motion.

The law states that the acceleration of an object is directly proportional to the net external force acting on it and inversely proportional to its inertial mass. In AP Physics 1, we almost always write it in vector form ${\vec{F}}_{\text{net}}=m\vec{a}$, where ${\vec{F}}_{\text{net}}$ is the vector sum of all external forces, $m$ is the inertial mass of the object, and $\vec{a}$ is the acceleration of the object's center of mass. Unlike common misconceptions, Newton's second law does not relate net force to velocity—only to acceleration. This means zero net force gives zero acceleration, not zero speed, which aligns with Newton's first law as a special case of the second law.

2. Vector Decomposition and 1-D Applications

Because ${\vec{F}}_{\text{net}}=m\vec{a}$ is a vector equation, it holds independently for each perpendicular component of force and acceleration. For any problem, the standard problem-solving workflow is: 1) draw a free-body diagram, 2) choose a coordinate system aligned with the direction of acceleration, 3) resolve all forces into x and y components, 4) sum the components to get net force in each direction, 5) set equal to mass times acceleration in that direction to solve for unknowns.

The most fundamental 1-D application is distinguishing mass and weight: mass is an invariant measure of an object's inertial resistance to acceleration (units: kg), while weight is the gravitational force acting on an object (units: N), given by $W=mg$, where $g=9.8{\text{ m/s}}^2$ is acceleration due to gravity near Earth's surface. This distinction is tested frequently on MCQs.

Worked Example

A 65 kg skydiver falls straight down, and experiences an upward drag force of 420 N. What is her acceleration?

1. Choose coordinate system with downward as the positive x direction (aligned with acceleration).
2. Identify all forces: downward weight $W=mg$, upward drag $F_d$.
3. Calculate net force: $F_{\text{net}}=W-F_d=(65)(9.8)-420=637-420=217\text{ N}$.
4. Rearrange Newton's second law to solve for acceleration: $a=F_{\text{net}}/m=217/65\approx 3.3{\text{ m/s}}^2$ downward.

Exam tip: Always choose your coordinate system so that the direction of acceleration lies along one of the axes. This eliminates cross terms, saves time, and drastically reduces sign errors on the exam.

3. Motion on Inclined Planes

Inclined plane problems are the most common 2-D Newton's second Law problem on AP Physics 1, as they test your ability to correctly decompose vectors and apply constraints to acceleration. The standard coordinate system for inclines aligns the x-axis parallel to the incline (direction of possible motion) and y-axis perpendicular to the incline.

Only weight is not aligned with the axes, so we decompose it into two components: parallel to the incline, $W_{\parallel }=mg\sin \theta$, and perpendicular to the incline, $W_{\perp }=mg\cos \theta$, where $\theta$ is the angle of the incline from the horizontal. For any fixed incline, an object cannot accelerate through the surface or jump off it unless explicitly stated, so $a_y=0$, meaning net force perpendicular to the incline is always zero. This lets us solve for normal force $N$ easily, which we then use to calculate friction if needed.

Worked Example

A 12 kg box slides down a frictionless 25° incline. Find the magnitude of the box's acceleration.

1. Align x parallel to the incline (positive downward), y perpendicular (positive outward from the surface).
2. Decompose weight: $W_x=mg\sin 25^{\circ }$, $W_y=-mg\cos 25^{\circ }$. Normal force $N=+N$ in the y-direction, no other forces act on the box.
3. Apply Newton's second law to the y-direction first: $F_{\text{net},y}=N-mg\cos 25^{\circ }=m{a}_y=0$, so $N=mg\cos 25^{\circ }$.
4. Apply Newton's second law to the x-direction: $F_{\text{net},x}=mg\sin 25^{\circ }=m{a}_x$. Mass cancels out, leaving $a=g\sin 25^{\circ }\approx 9.8(0.42)\approx 4.1{\text{ m/s}}^2$.

Exam tip: To avoid swapping sine and cosine for weight components, check the edge case $\theta =90^{\circ }$ (vertical incline = free fall). At this angle, $a=g$, which only works if the parallel component uses sine ($\sin 90^{\circ }=1$).

4. Connected Objects and System Approach

When multiple objects are connected by a taut massless string over a massless, frictionless pulley, they move with the same magnitude of acceleration, so we can solve the system two ways: isolate each object and solve a system of equations, or treat the entire connected group as a single system.

For the system approach, internal forces (like tension in the connecting string between the objects) cancel out by Newton's third law, so we only need to include external forces when calculating net force. This drastically simplifies finding the acceleration of the system, but if we need to find the tension force itself, we still have to isolate one of the objects to solve for it, since tension is internal to the full system.

Worked Example

A 3.0 kg block on a horizontal frictionless table is connected by a massless string over a massless pulley to a 1.5 kg hanging block. When released, what is the acceleration of the blocks?

1. Confirm the magnitude of acceleration is the same for both blocks, so the system approach is valid.
2. Mark the system boundary around both blocks: tension is internal, so it cancels. The only external net force along the direction of motion is the weight of the hanging block (weight and normal force on the table block cancel each other out).
3. Calculate total mass of the system: $M_{\text{total}}=3.0+1.5=4.5\text{ kg}$. Net force: $F_{\text{net}}=m_{\text{hanging}}g=1.5(9.8)=14.7\text{ N}$.
4. Apply Newton's second law to the system: $a=F_{\text{net}}/M_{\text{total}}=14.7/4.5\approx 3.3{\text{ m/s}}^2$, directed toward the hanging block.

Exam tip: If you need tension, always isolate one object after finding acceleration. For this example, isolating the table block gives $T=m_{\text{table}}a=3.0(3.3)=9.9\text{ N}$, which is correct.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Writing $a=v/F_{net}$ instead of $a=F_{net}/m$, or rearranging incorrectly for unknown variables. Why: Confuses the proportionality relationships by memorizing rearranged forms instead of starting from the original law. Correct move: Always write ${\vec{F}}_{net}=m\vec{a}$ at the top of your solution, then rearrange step-by-step for whatever variable you need.
• Wrong move: Decomposing weight on an incline as $W_{\parallel }=mg\cos \theta$ and $W_{\perp }=mg\sin \theta$. Why: Swaps sine and cosine by mixing up the angle in the weight decomposition right triangle. Correct move: Always check with the $\theta =0^{\circ }$ (flat ground) edge case: parallel acceleration should be zero, so $\sin 0^{\circ }=0$ confirms sine belongs to the parallel component.
• Wrong move: Using weight directly (in Newtons) in place of mass in $F_{net}=ma$ when a problem gives an object's weight instead of mass. Why: Confuses mass (inertia, kg) and weight (force, N), even when the problem explicitly gives weight. Correct move: If a problem states a "50 N crate", first calculate mass from $m=W/g$ before plugging into Newton's second law.
• Wrong move: Including tension between connected objects when calculating net force for the full system. Why: Forgets that internal action-reaction pairs cancel out, so they do not contribute to the system's acceleration. Correct move: Before calculating net force for a system, cross out any force that acts between two objects both inside your system boundary.
• Wrong move: Assigns positive signs to both the normal force and the perpendicular component of weight on an incline, leading to a non-zero acceleration perpendicular to the surface. Why: Poor sign convention aligned to the coordinate system. Correct move: Always draw your coordinate system on the free-body diagram, then assign a sign to every force based on whether it points along the positive or negative axis before summing.
• Wrong move: Setting net force perpendicular to a fixed incline equal to $ma$. Why: Applies Newton's second law uniformly without accounting for the physical constraint of the incline. Correct move: For any fixed incline, explicitly note that $a_{\perp }=0$, so net force perpendicular to the incline is zero.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

An experiment is done to test Newton's second law: a cart on a frictionless track is pulled by different hanging masses, and the net force on the cart and the cart's acceleration are measured for multiple trials. If the student plots acceleration on the y-axis and net force on the x-axis, what is the physical meaning of the slope of the resulting linear graph? A) The mass of the cart B) The reciprocal of the mass of the cart C) The weight of the cart D) The acceleration due to gravity

Worked Solution: Start with Newton's second law $F_{\text{net}}=ma$, and rearrange it to match the form of the plotted graph: $a=\left(\frac{1}{m}\right)F_{\text{net}}$. This matches the linear equation $y=mx+b$, where $y=a$ (y-axis), $x=F_{\text{net}}$ (x-axis), and the slope is equal to $\frac{1}{m}$, the reciprocal of the cart's mass. The y-intercept is zero, which matches this relationship. The correct answer is B.

Question 2 (Free Response)

A 5.0 kg crate is pulled up a 20° rough incline by a rope with tension 40 N parallel to the incline. The coefficient of kinetic friction between the crate and the incline is 0.30. (a) Draw a free-body diagram for the crate, labeling all forces. (b) Calculate the normal force exerted on the crate by the incline. (c) Calculate the acceleration of the crate. (d) Is the crate speeding up or slowing down as it moves up the incline? Justify your answer.

Worked Solution: (a) Four forces labeled on the free-body diagram: (1) Weight $W$ pointing vertically downward, (2) Normal force $N$ pointing perpendicular to the incline outward, (3) Tension $T$ pointing up the incline parallel to the surface, (4) Kinetic friction $f_k$ pointing down the incline (opposite the direction of motion). (b) Align y perpendicular to the incline, with positive outward. There is no acceleration perpendicular to the incline, so: $$F_{\text{net},y}=N-mg\cos \theta =0$$ $$N=(5.0)(9.8)\cos (20^{\circ })\approx 49(0.94)=\boxed{46\text{ N}}$$ (c) Align x parallel to the incline, positive up. First calculate kinetic friction: $f_k={\mu }_{k}N=0.30(46)=13.8\text{ N}$. Net force is: $$F_{\text{net},x}=T-mg\sin \theta -f_k=40-(49)(0.34)-13.8=9.5\text{ N}$$ $$a=\frac{F_{\text{net},x}}{m}=\frac{9.5}{5.0}=\boxed{1.9{\text{ m/s}}^2\text{ up the incline}}$$ (d) Acceleration points in the same direction as the velocity (up the incline), so the crate is speeding up.

Question 3 (Application / Real-World Style)

A 1500 kg elevator accelerates upward uniformly from rest to 4.0 m/s in 8.0 s when moving to a higher floor. What is the tension in the elevator cable during this acceleration, which supports the entire mass of the elevator? Interpret your answer relative to the cable tension when the elevator is stationary.

Worked Solution: First find acceleration from kinematics: $a=\frac{\Delta v}{\Delta t}=\frac{4.0-0}{8.0}=0.50{\text{ m/s}}^2$ upward. Choose upward as positive, so net force gives: $$F_{\text{net}}=T-mg=ma$$ Rearrange for tension: $T=m(g+a)=1500(9.8+0.50)=15450\text{ N}$. When stationary, $a=0$, so tension is $T_0=mg=14700\text{ N}$. The tension during upward acceleration is 750 N (≈5%) larger than the tension when the elevator is stationary, which matches the common experience of feeling heavier when an elevator first starts moving upward.

7. Quick Reference Cheatsheet

8. What's Next

Newton's second law is the foundational quantitative tool for all of classical mechanics, so mastering it is required for every topic that comes after it in the AP Physics 1 syllabus. Next, you will apply Newton's second law to circular motion, where we relate centripetal net force to centripetal acceleration to solve problems involving orbits, banked curves, and roller coasters. Without a solid understanding of how to calculate net force and connect it to acceleration from Newton's second law, you will not be able to correctly analyze circular motion or later topics like energy conservation and momentum, which also build on force concepts. It also connects directly to the study of simple harmonic motion, where we relate net restoring force to acceleration of oscillating objects.

Free-Body Diagram Analysis Uniform Circular Motion Work and Energy Momentum Conservation