How the MCAT tests kinematics and forces

This material anchors AAMC content category 4A — translational motion, forces, work, and energy in living systems. The MCAT emphasizes setting up problems correctly: choosing the right kinematic equation, drawing a clean free-body diagram, and resolving vectors. Careful setup beats memorized formulas.

workspace_premium The universal first step

Almost every mechanics problem starts the same way: draw the free-body diagram, pick axes, and apply F = ma along each axis. Set that up correctly and the algebra follows.

The kinematic equations

For constant acceleration, four equations connect displacement (x), initial and final velocity (v₀, v), acceleration (a), and time (t). Pick the one that omits the variable you neither know nor want.

v = v₀ + atNo displacementRelates velocity and time directly through acceleration.
x = v₀t + ½at²No final velocityGives displacement from initial velocity, acceleration, and time.
v² = v₀² + 2axNo timeThe go-to when time is neither given nor asked for.

Projectile motion

A projectile moves under gravity alone. The key insight is that the horizontal and vertical motions are completely independent and share only the time.

vₓ v_y max height (v_y = 0) range vₓ constant; v_y changes with gravity
Projectile motion. Horizontal velocity vₓ stays constant; vertical velocity v_y decreases going up, is zero at the apex, and increases coming down.
  • Horizontal: no acceleration, so x = vₓt (constant velocity).
  • Vertical: acceleration = −g, so treat it as free fall.
  • At the peak, v_y = 0 but vₓ is unchanged, so the object is still moving horizontally.

Newton’s laws of motion

Newton’s three laws connect force to motion and are the backbone of every mechanics problem.

  1. First law (inertia): an object keeps its velocity unless a net force acts on it.
  2. Second law: F_net = ma — net force produces acceleration in the same direction.
  3. Third law: forces come in equal and opposite pairs acting on different objects.
F_net = maNewton’s second lawSum the forces along each axis and set equal to mass times acceleration. In equilibrium, a = 0 and forces balance.

Free-body diagrams and inclined planes

A free-body diagram isolates one object and draws every force acting on it. On an inclined plane, tilt your axes to align with the ramp so gravity is the only force you have to resolve.

mg N f θ weight splits intomg·sinθ (down ramp)mg·cosθ (into ramp)
Block on an incline. Weight (mg) points straight down; the normal force (N) is perpendicular to the surface; friction (f) acts along it. Gravity resolves into mg·sinθ down the ramp and mg·cosθ into it.
functions Worked example — acceleration down a frictionless incline

A block slides down a frictionless 30° incline. What is its acceleration? (g = 10 m/s²)

  1. Along the ramp, the only force is the gravity component mg·sinθ.
  2. Newton’s second law along the ramp: ma = mg·sinθ, so a = g·sinθ.
  3. a = 10 · sin30° = 10 · 0.5 = 5 m/s².

Answer. a = 5 m/s² down the ramp — mass cancels, so it does not matter.

Friction, tension, and the normal force

Three contact forces recur constantly. Getting their directions and magnitudes right is most of the battle.

area = displacement slope = acceleration Time → Velocity →
Reading motion graphs. On a velocity–time graph, the slope is acceleration and the area under the curve is displacement.
  • Normal force (N) is perpendicular to the surface and adjusts to prevent objects from passing through it — it is not always equal to mg.
  • Friction (f = μN) opposes relative motion; static friction can range up to μₛN, while kinetic friction stays about μₖN.
  • Tension pulls equally along a massless rope and is the same throughout an ideal string over a frictionless pulley.

How to study it — and the traps to avoid

Mechanics rewards disciplined setup. Practice the same routine on every problem until it is reflexive.

  1. Draw a free-body diagram before writing any equation.
  2. Separate motion into independent x and y components (especially for projectiles).
  3. On inclines, rotate axes to align with the ramp.
  4. Check whether the system is in equilibrium (a = 0) or accelerating.
error Highest-yield trap

The normal force does not always equal mg. On an incline it is mg·cosθ, and in an accelerating elevator it changes — that is exactly why you feel heavier or lighter. Always get N from the free-body diagram, not by assuming N = mg.