Newton's Second Law — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: vector form of Newton's second law, mass vs weight distinction, inertial reference frames, system vs particle approaches, connected objects analysis, and application of $F=ma$ to constant and variable acceleration problems tested on both sections of the exam.

You should already know: Vector decomposition and coordinate system setup for motion in multiple dimensions. Definition of force and inertia from Newton's first law. Basic kinematic relationships between position, velocity, and acceleration.

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 C: Mechanics 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 foundational relationship between net force, mass, and acceleration that underpins almost all of classical mechanics, making up ~12-18% of the total exam weight for AP Physics C: Mechanics (per the College Board CED for Unit 2: Newton's Laws of Motion). It appears in every section of the exam: standalone multiple-choice questions, embedded conceptual questions in FRQs, and as a core required step in almost every free-response problem involving forces or motion. The formal definition states that 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 the familiar ${\vec{F}}_{net}=m\vec{a}$, where ${\vec{F}}_{net}$ is the vector sum of all external forces, $m$ is the inertial mass of the body, and $\vec{a}$ is the acceleration of the body's center of mass. Unlike common misconceptions, force does not cause velocity—it causes a change in velocity, i.e., acceleration. This topic is the bridge between the kinematics you learned in Unit 1 and the dynamic analysis of motion that is the core of AP Physics C.

2. Vector Decomposition and Inertial Reference Frames

The vector nature of Newton's second law means we can break ${\vec{F}}_{net}$ and $\vec{a}$ into components along any orthogonal coordinate system we choose, so we can write separate, independent equations for each axis: $F_{net,x}=m{a}_x$, $F_{net,y}=m{a}_y$, $F_{net,z}=m{a}_z$. This is the first step in almost every problem involving forces on the exam, and choosing a convenient coordinate system can cut your work in half.

A critical requirement for Newton's second law to hold is that we work in an inertial reference frame: any frame that is not accelerating relative to the fixed stars. For all AP problems, the ground (or any surface fixed to the Earth) is an excellent approximation of an inertial frame, and we can ignore the tiny centripetal acceleration from Earth's rotation. Non-inertial (accelerating) frames introduce fictitious forces that are not real interactions between objects, so we always convert to an inertial frame to solve problems on the exam.

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 sets acceleration to be non-zero only along the $x$-axis, simplifying calculations.
2. Decompose the weight $mg$ into components: $m{g}_x=mg\sin \theta$ (down the incline), $m{g}_y=-mg\cos \theta$ (into the incline). Normal force $N$ points along $+y$, kinetic friction $f_k={\mu }_{k}N$ points along $-x$ (opposing motion).
3. Apply Newton's second law to the $y$-axis: acceleration perpendicular to the incline is zero, so $F_{net,y}=N-mg\cos \theta =0⟹N=mg\cos \theta$.
4. Apply Newton's second law to the $x$-axis: $F_{net,x}=mg\sin \theta -{\mu }_{k}N=ma$. Substitute $N=mg\cos \theta$ to get $a=g(\sin \theta -{\mu }_k\cos \theta )$.
5. Plug in values ($g=9.8{\text{m/s}}^2$): $a=9.8(0.5-0.2\cdot \frac{\sqrt{3}}{2})\approx 3.2{\text{m/s}}^2$ down the incline.

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. System vs Particle Approach for Connected Objects

When solving problems with multiple connected objects (e.g., blocks connected by ropes, modified Atwood machines), you can use two valid approaches: treating each object as an individual particle, or treating all connected objects as a single system. When treating as a system, only external forces contribute to ${\vec{F}}_{net}$; internal forces (tension between blocks, contact forces between objects) cancel out by Newton's third law, so you can ignore them entirely when calculating the acceleration of the whole system.

The particle approach is required if you need to find the value of an internal force (e.g., tension in a rope between two blocks), while the system approach can quickly give acceleration without solving a system of equations, saving time on multiple-choice questions. Always match your approach to what the question is asking for to avoid unnecessary work.

Worked Example

Two blocks of mass $m_1=2\text{kg}$ and $m_2=3\text{kg}$ 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. First, use the system approach to find the acceleration of the whole system. The only external force causing acceleration is the weight of $m_2$, so $F_{net}=m_{2}g=(m_1+m_2)a$. Solve for $a$: $a=\frac{m_{2}g}{m_1+m_2}=\frac{3(9.8)}{5}=5.88{\text{m/s}}^2$.
2. Tension is an internal force of the system, so we switch to the particle approach applied to $m_1$, where tension is the only horizontal force acting on the block.
3. Apply Newton's second law to $m_1$: $T=m_{1}a=m_1\left(\frac{m_{2}g}{m_1+m_2}\right)=\frac{2\cdot 3\cdot 9.8}{5}=11.76\text{N}\approx 11.8\text{N}$.
4. Verify by applying Newton's second law to $m_2$: $m_{2}g-T=m_{2}a⟹T=3(9.8-5.88)=11.76\text{N}$, which matches our result.

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

Unlike algebra-based physics, AP Physics C: Mechanics regularly tests application of Newton's second law to variable forces, which requires calculus. The general form of Newton's second law, valid for any system, is: $${\vec{F}}_{net}=\frac{d\vec{p}}{dt}=\frac{d(m\vec{v})}{dt}$$ For constant mass systems, this simplifies to ${\vec{F}}_{net}=m\frac{d\vec{v}}{dt}=m\vec{a}$, which matches our earlier formula. When force varies as a function of time, position, or velocity, we can rearrange this relationship to integrate and solve for velocity or position as a function of time. This is one of the most common calculus-based applications of Newton's laws on the AP exam.

Worked Example

A 2 kg object moving along the x-axis is acted on by a time-varying force $F(t)=4t-2t^2$ (in Newtons, for $t\ge 0$). At $t=0$, the object has an initial velocity of $v_0=3\text{m/s}$. Find the velocity of the object at $t=2\text{s}$.

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

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. Common Pitfalls (and how to avoid them)

• 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 $F_{net}$ 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.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

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 $\sin 37^{\circ }=0.6$, $\cos 37^{\circ }=0.8$, $g=9.8{\text{m/s}}^2$) A) 49 N B) 43 N C) 55 N D) 39 N

Worked Solution: First, draw a free-body diagram for the block: forces are downward weight, upward normal force, and the applied force at 37° above horizontal. Acceleration along the vertical axis is zero, so net vertical force must equal zero by Newton's second law. Summing vertical forces gives $N+F\sin \theta -mg=0$. Calculate $mg=5(9.8)=49$ N, then rearrange to get $N=mg-F\sin \theta =49-10(0.6)=43$ N. The horizontal component of the force affects acceleration but not normal force, so options A (ignoring the vertical component of the applied force) and C (using the wrong sign) are common errors. The correct answer is B.

Question 2 (Free Response)

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 $2M$. (a) Draw free-body diagrams for each block, and derive an expression for the magnitude of the acceleration of the blocks in terms of $M$ and $g$. (b) Derive an expression for the tension $T$ in the string in terms of $M$ and $g$. (c) If $M=1.5\text{kg}$, what is the magnitude of the net force on the 2M block?

Worked Solution: (a) Free-body diagrams: each block has a downward gravitational force ($Mg$ for block 1, $2Mg$ for block 2) and an upward tension force $T$ (equal magnitude for both, since the string is massless and pulley is massless/frictionless). Let $a$ be the magnitude of acceleration, upward for $M$ and downward for $2M$. Newton's second law gives: $T-Mg=Ma$ (for $M$) $2Mg-T=2Ma$ (for $2M$) Add the two equations to eliminate $T$: $Mg=3Ma⟹a=\frac{g}{3}$.

(b) Substitute $a=g/3$ back into the first equation: $T=Mg+Ma=Mg+M(g/3)=\frac{4}{3}Mg$. Verifying with the second equation gives the same result.

(c) By Newton's second law, net force on the 2M block equals mass times acceleration: $F_{net}=(2M)a=2M(g/3)=2(1.5)(9.8/3)=9.8\text{N}$.

Question 3 (Application / Real-World Style)

A 1500 kg empty elevator accelerates upward at $1.2{\text{m/s}}^2$ to pick up a passenger at the 5th floor. The elevator cable can withstand a maximum tension of 20,000 N before breaking. Does the cable break under these conditions? Justify your answer with calculations.

Worked Solution: Draw a free-body diagram for the elevator: upward tension $T$, downward weight $mg$. Take upward as the positive direction, and apply Newton's second law: $T-mg=ma⟹T=m(g+a)$. Substitute values: $T=1500(9.8+1.2)=1500(11)=16,500\text{N}$. This tension is 3,500 N lower than the maximum 20,000 N the cable can withstand. In context, the cable has sufficient margin of safety and does not break under this acceleration for an empty elevator.

7. Quick Reference Cheatsheet

8. 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 chapter, solving circular motion problems will be extremely difficult. Later, Newton's second law is the starting point for work and energy, rotational dynamics, and oscillations—every major unit after this builds on the $F_{net}=ma$ relationship. This topic 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.

Circular Motion Work and Kinetic Energy Rotational Dynamics Simple Harmonic Motion