What is AP Physics C: Mechanics unit 2?
Force and Translational Dynamics is the largest conceptual foundation in AP Physics C: Mechanics. Every later unit on energy, momentum, rotation, and oscillations depends on the force models and problem-solving strategies introduced here.
Unit 2 teaches you how to identify all forces on a system, represent them in a free-body diagram, and apply Newton's laws to predict motion. It also covers specific force laws: universal gravitation, friction, Hooke's law, linear drag, and the centripetal force requirement for circular paths.
Systems and Newton's Laws
Topics 2.1-2.5 establish how to define a system, locate its center of mass, draw a correct free-body diagram, and apply Newton's first and second laws. The center of mass equation for discrete masses is x_cm = (sum of m_i x_i) / (sum of m_i), and for continuous objects you integrate r_cm = integral of r dm / integral of dm. Newton's second law gives a_sys = F_net / m_sys.
Specific Force Models
Topics 2.6-2.9 introduce four force laws you must apply quantitatively. Universal gravitation: F_g = G m1 m2 / r^2. Friction: kinetic F_fk = mu_k F_N; static F_fs is less than or equal to mu_s F_N. Hooke's law: F_s = -k delta x. Linear drag: F_r = -k v, which produces exponential velocity approaching terminal speed.
Circular Motion and Orbits
Topic 2.10 requires identifying which real forces provide the net inward (centripetal) force. Centripetal acceleration is a_c = v^2 / r, always directed toward the center. For satellites in circular orbit, gravity supplies all centripetal force, and Kepler's third law gives T^2 = (4 pi^2 / GM) R^3.
Forces as interactions, not propertiesEvery force in this unit is an interaction between two objects. Newton's third law guarantees a paired force on the other object, internal forces never change a system's center-of-mass motion, and the net external force alone determines acceleration. This interaction-based view is the thread connecting free-body diagrams, gravity, friction, springs, drag, and circular motion throughout the unit.
Unit 2 review notes
2.1
Systems and Center of Mass
A system is a collection of objects whose interactions you choose to analyze together. If internal structure does not matter, you can treat the entire system as a point mass located at its center of mass. For discrete masses, x_cm = (sum of m_i x_i) / (sum of m_i). For a continuous object with variable density, r_cm = integral of r dm / integral of dm, where dm = lambda d-ell for a rod with linear mass density lambda = dm / d-ell. Symmetry lines always pass through the center of mass of a uniform object.
- System boundary: Defines which objects are internal; only external forces change the center-of-mass motion.
- Center of mass (discrete): x_cm = (sum m_i x_i) / (sum m_i); mass-weighted average position of all constituent objects.
- Center of mass (continuous): r_cm = integral of r dm / integral of dm; requires expressing dm in terms of a density function and a position variable.
- Linear mass density: lambda = dm / d-ell; used to set up the dm integral for rods or wires with nonuniform density.
- Symmetry argument: For any uniform object with a line of symmetry, the center of mass lies on that line, often eliminating the need for integration.
A nonuniform rod of length L has linear mass density lambda(x) = 2x kg/m. Set up the integral to find x_cm and identify the limits of integration.
| Mass distribution type | dm expression | Integration variable |
|---|
| Uniform rod | lambda dx (constant lambda) | x from 0 to L |
| Nonuniform rod | lambda(x) dx | x from 0 to L |
| Uniform disk (radial) | sigma 2 pi r dr | r from 0 to R |
2.2
Forces and Free-Body Diagrams
A force is a vector interaction between two objects; an object cannot exert a net force on itself. A free-body diagram shows every external force on a single object as an arrow originating from a dot representing the center of mass. Draw one arrow per force, label each with its type (weight, normal, tension, friction), and choose a coordinate axis aligned with the direction of acceleration to simplify the algebra. Do not include force components as separate arrows on the diagram.
- Contact force: Arises when two objects touch; macroscopic result of interatomic electric forces. Examples: normal force, friction, tension.
- Free-body diagram rule: Each force arrow starts at the center-of-mass dot and points in the direction the force acts; components are written in equations, not drawn as extra arrows.
- Axis choice: Rotating the coordinate system so one axis is parallel to the acceleration reduces the number of simultaneous equations needed.
- Net force: Vector sum of all external forces; equals m times a by Newton's second law.
Draw a free-body diagram for a block sliding down a rough incline. Identify every force and explain your axis choice.
2.3
Newton's Third Law and Tension
Newton's third law states F_A on B = -F_B on A. The paired forces are equal in magnitude, opposite in direction, and act on different objects, so they never cancel in a single free-body diagram. Internal forces between objects within a system do not affect the system's center-of-mass acceleration. Tension in an ideal (massless, inextensible) string is the same at every point; in a string with nonnegligible mass, tension varies along the string. An ideal pulley has negligible mass and frictionless axle, so it only redirects tension without changing its magnitude.
- Action-reaction pair: Two forces that are equal in magnitude and opposite in direction, each acting on a different object in the interaction.
- Internal forces: Forces between objects within the chosen system; they cancel in pairs and do not change the system's center-of-mass motion.
- Tension (ideal string): Uniform throughout a massless, inextensible string; the string transmits force without storing it.
- Ideal pulley: Massless and frictionless; redirects tension so its magnitude is the same on both sides of the pulley.
Two blocks connected by an ideal string are pulled across a frictionless surface. Identify one Newton's third law pair and explain why the string tension is the same throughout.
| String type | Tension uniform? | Effect on analysis |
|---|
| Ideal (massless, inextensible) | Yes | Single tension value T throughout |
| Real (nonnegligible mass) | No | Tension varies; must treat string as extended object |
2.4
Newton's First and Second Laws
Newton's first law: if the net force on a system is zero, its velocity is constant (translational equilibrium, sum of F_i = 0). Forces can be balanced along one axis and unbalanced along another; only the unbalanced direction produces acceleration. An inertial reference frame is one in which Newton's first law holds. Newton's second law: a_sys = F_net / m_sys. The acceleration of the center of mass points in the same direction as the net external force. Apply it component by component after drawing a free-body diagram.
- Translational equilibrium: Net force equals zero in every direction; the object moves at constant velocity or stays at rest.
- Inertial reference frame: A frame in which an object with zero net force moves at constant velocity; Newton's laws hold without fictitious forces.
- Newton's second law: a_sys = F_net / m_sys; acceleration is proportional to net external force and inversely proportional to mass.
- Component form: Apply sum of F_x = m a_x and sum of F_y = m a_y separately after choosing a coordinate system aligned with the acceleration.
A 5 kg block on a frictionless surface is pulled by two horizontal forces: 20 N east and 8 N west. Find the acceleration magnitude and direction.
| Condition | Net force | Motion result |
|---|
| Translational equilibrium | Zero in all directions | Constant velocity (or rest) |
| Unbalanced in one direction | Nonzero along one axis | Acceleration along that axis only |
| Unbalanced in two directions | Nonzero along both axes | Acceleration with both components |
2.6
Gravitational Force
Newton's law of universal gravitation: F_g = G m1 m2 / r^2, attractive, along the line connecting the two centers of mass. Near Earth's surface, g is approximately 10 N/kg and gravitational force is treated as constant (weight = mg). The gravitational field at a point is g = G M / r^2. Apparent weight equals the normal force; it differs from true weight when the system accelerates. The equivalence principle states that an observer in a noninertial frame cannot distinguish apparent weight from gravitational weight. Newton's shell theorem: a uniform spherical shell exerts zero net gravitational force on an object inside it, and acts as a point mass at its center for objects outside. An object inside a uniform solid sphere feels only the gravitational pull from the partial mass enclosed within its radius.
- Universal gravitation: F_g = G m1 m2 / r^2; force is attractive, acts along the line of centers, and follows an inverse-square law.
- Gravitational field: g = G M / r^2 at distance r from mass M; units are N/kg, numerically equal to free-fall acceleration.
- Apparent weight: Magnitude of the normal force on a system; equals mg only when acceleration is zero.
- Newton's shell theorem: Inside a uniform spherical shell: zero net force. Outside: shell acts as a point mass at its center.
- Equivalence principle: Inertial mass and gravitational mass are experimentally equal; an accelerating observer cannot distinguish their apparent weight from a gravitational field.
An object is located at radius r inside a uniform solid sphere of mass M and radius R. Write an expression for the gravitational force on the object in terms of r, R, M, and G.
| Object location | Net gravitational force from shell/sphere |
|---|
| Inside thin spherical shell | Zero |
| Outside thin spherical shell | G M_shell / r^2 (shell acts as point mass) |
| Inside uniform solid sphere at radius r | G M_partial / r^2, where M_partial = M(r/R)^3 |
2.7
Kinetic and Static Friction
Kinetic friction acts when two surfaces slide relative to each other: F_fk = mu_k F_N, directed opposite to the relative motion. Static friction acts when surfaces are not sliding; it adjusts in magnitude and direction to prevent motion up to a maximum of F_fs,max = mu_s F_N. Because mu_s is typically greater than mu_k, more force is needed to start sliding than to maintain it. Friction magnitude does not depend on contact area. Normal force is perpendicular to the surface, not always equal to mg (for example, on an incline or when an additional vertical force is applied).
- Kinetic friction: F_fk = mu_k F_N; constant magnitude once sliding begins, directed opposite to relative motion.
- Static friction: F_fs is less than or equal to mu_s F_N; self-adjusting force that prevents sliding until the applied force exceeds the maximum.
- Normal force: Perpendicular contact force from the surface; determines friction magnitude but is not always equal to mg.
- mu_s greater than mu_k: The coefficient of static friction exceeds kinetic friction for the same surfaces, so objects require more force to start moving than to keep moving.
A 10 kg block sits on a surface with mu_s = 0.5 and mu_k = 0.3. What is the minimum horizontal force needed to start the block moving, and what is the friction force once it is sliding at constant velocity?
| Friction type | Equation | When it applies |
|---|
| Static | F_fs is less than or equal to mu_s F_N | Surfaces not moving relative to each other |
| Kinetic | F_fk = mu_k F_N | Surfaces sliding relative to each other |
2.8
Spring Forces
An ideal spring has negligible mass and obeys Hooke's law: F_s = -k delta x, where delta x is displacement from the relaxed length and k is the spring constant in N/m. The negative sign means the force always points back toward equilibrium. For springs in series, 1/k_eq = sum of 1/k_i, giving an equivalent constant smaller than any individual spring. For springs in parallel, k_eq = sum of k_i, giving a stiffer combined spring. A nonideal spring either has nonnegligible mass or a force that is not proportional to displacement.
- Hooke's law: F_s = -k delta x; restoring force proportional to displacement from equilibrium, directed toward equilibrium.
- Spring constant k: Stiffness of the spring in N/m; larger k means more force per unit displacement.
- Springs in series: 1/k_eq = 1/k1 + 1/k2 + ...; equivalent constant is smaller than the smallest individual k.
- Springs in parallel: k_eq = k1 + k2 + ...; equivalent constant is the sum of all individual constants.
Two springs with k1 = 200 N/m and k2 = 100 N/m are connected in series. Find k_eq and compare it to each individual spring constant.
| Configuration | k_eq formula | k_eq relative to components |
|---|
| Series | 1/k_eq = 1/k1 + 1/k2 + ... | Smaller than smallest k |
| Parallel | k_eq = k1 + k2 + ... | Larger than largest k |
2.9
Resistive Forces
A linear resistive (drag) force is modeled as F_r = -k v, opposing the velocity. Applying Newton's second law gives m dv/dt = F_applied - k v, a first-order separable differential equation. Solving by separation of variables with initial condition v(0) = v0 yields v(t) = v_terminal + (v0 - v_terminal) e^(-kt/m), where v_terminal = F_applied / k and the time constant is tau = m/k. As t increases, velocity approaches terminal speed exponentially. Acceleration and position are also exponential functions of time. For a falling object, v_terminal = mg/k.
- Linear drag force: F_r = -k v; magnitude proportional to speed, direction always opposite to velocity.
- Separation of variables: Technique for solving m dv/dt = F - kv by rearranging to dv/(F - kv) = dt/m and integrating both sides.
- Terminal speed: v_terminal = F_applied / k; the constant speed at which drag equals the driving force and acceleration reaches zero.
- Time constant tau: tau = m/k; the time for the velocity difference from terminal speed to decrease by a factor of e.
A 2 kg object falls from rest with drag coefficient k = 4 N/(m/s). Write the differential equation, identify v_terminal, and write v(t).
2.10
Circular Motion and Orbits
An object in circular motion has centripetal acceleration a_c = v^2 / r directed toward the center. The net inward force from all real forces (gravity, normal, tension, friction) must equal m v^2 / r. For a vertical loop, the minimum speed at the top occurs when gravity alone provides centripetal force: v_min = sqrt(g r). For a conical pendulum, the horizontal component of tension provides centripetal force and the vertical component balances gravity. For a satellite in circular orbit, gravity is the only centripetal force, giving Kepler's third law: T^2 = (4 pi^2 / GM) R^3.
- Centripetal acceleration: a_c = v^2 / r; always directed toward the center of the circular path, not a separate force but the result of the net inward force.
- Net centripetal force: Sum of inward force components from all real forces equals m v^2 / r; identify which forces point inward and which point outward.
- Minimum speed at loop top: v_min = sqrt(g r); at this speed, normal force is zero and gravity alone provides centripetal acceleration.
- Kepler's third law: T^2 = (4 pi^2 / GM) R^3 for a circular orbit; relates orbital period to orbital radius and the mass of the central body.
- Tangential acceleration: Rate of change of speed along the path; nonzero in nonuniform circular motion, perpendicular to centripetal acceleration.
A car travels over a hill of radius r at speed v. Write the Newton's second law equation at the top of the hill and find the speed at which the car loses contact with the road.
| Scenario | Force providing centripetal acceleration | Key equation |
|---|
| Horizontal circle (string) | Tension component inward | T = m v^2 / r |
| Vertical loop (top) | Gravity + normal force inward | mg + N = m v^2 / r |
| Conical pendulum | Horizontal tension component | T sin(theta) = m v^2 / r |
| Satellite orbit | Gravity only | G M m / R^2 = m v^2 / R |
Practice AP Physics C: Mechanics unit 2 questions
Try AP-style multiple-choice questions and written prompts after you review the notes.
QuestionA student plots the square of the orbital period T2 on the vertical axis versus the cube of the orbital radius r3 on the horizontal axis for Jupiter's moons. The best-fit line has a slope S. Which expression correctly determines Jupiter's mass MJ using this slope?
MJ=GS4π2, because the slope represents the constant GMJ4π2.
MJ=4π2GS, because the slope represents the constant 4π2GMJ.
MJ=4π2GS, because the slope represents the constant 4π2GMJ.
MJ=G4π2S, because the slope represents the constant 4π2MJG.
QuestionA crate of mass m is pulled across a rough horizontal floor by a rope angled θ above the horizontal with tension T. Which claim correctly describes how the kinetic friction force fk depends on the angle θ (for 0<θ<90∘)?
It decreases as θ increases because the vertical tension component reduces the normal force
It increases as θ increases because the horizontal tension component increases the speed
It remains constant as θ increases because the normal force equals the crate's weight
It decreases as θ increases because the coefficient of friction varies with the angle
2. A small block of mass m = 0.50 kg is placed on a horizontal turntable at a distance r = 0.40 m from the center of the turntable, as shown in Figure 1. The turntable rotates about a vertical axis through its center. The coefficient of static friction between the block and the turntable is μₛ = 0.60, and the coefficient of kinetic friction is μₖ = 0.40.
Figure 1: Top view of turntable with block at radius r = 0.40 m (counterclockwise rotation)
Figure 2: Dot represents the block in Scenario 1 (viewed from above)
Figure 3: Axes for graphing friction force magnitude f versus angular speed ω
4. Two identical wooden blocks, each of mass m = 2.0 kg, are placed on two different inclined planes, as shown in Figure 1 and Figure 2. Both planes make an angle θ = 30° with the horizontal. The coefficient of kinetic friction between each block and its respective surface is μk = 0.20.
Indicate whether a₁ is greater than, less than, or equal to a₂ by writing one of the following.
Justify your answer using qualitative reasoning beyond referencing equations.
Indicate whether the distance d₃ that the new block travels up the incline before coming to rest is greater than, less than, or equal to the distance d₁ from Scenario 1.
Briefly justify your answer.
1. A block of mass M is placed on a horizontal surface with coefficient of kinetic friction μ_k. One end of an ideal spring with spring constant k is attached to the block, and the other end is attached to a fixed wall. The block is initially at position x = -x_0, where the spring is compressed by a distance x_0 from its relaxed length. The relaxed position of the spring corresponds to x = 0. The block is released from rest at x = -x_0 and slides along the surface in the +x-direction, as shown in Figure 1. The block reaches position x = d before coming to rest.
Figure 1. Block–spring system on a rough horizontal surface. The block is released from rest at x = −x₀ (spring compressed) and later comes to rest at x = d.
Figure 2. Draw and label the forces acting on the block while it slides on the horizontal surface between x = −x₀ and x = 0.
i. On the dot in Figure 2, draw and label arrows to represent all the forces exerted on the block at an instant while it is sliding between x = -x_0 and x = 0. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.
ii. The block slides from x = -x_0 to x = d, where it comes to rest. During this motion, the force exerted on the block by the spring varies with position according to F_s(x) = -kx, where k is the spring constant.
Derive an expression for the distance d in terms of x_0, μ_k, M, g, and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.