Newton's Second Law

Newton's Second Law

9 min read Updated Mar 26, 2026

A nurse pushes two hospital beds down the same hallway with the same effort. The empty bed glides easily; the bed carrying a 200 kg patient barely budges. Same push. Different result. Why?

That’s Newton’s second law in one sentence: the same force produces less acceleration on more mass. Push harder, you go faster. Push the same thing that’s heavier, you go slower. It’s the rule that connects “what you do” (force) to “what happens” (acceleration), and it’s the single most-used equation on MCAT physics.

The Law

In plain English, this equation says three things:

  1. Push harder, accelerate faster. Twice the force on the same object → twice the acceleration.
  2. Heavier things are harder to speed up. Twice the mass with the same force → half the acceleration.
  3. Acceleration points where the net force points. Not where the object is going — where the forces are pulling it.

That third point is the one MCAT loves to trick you on. A car braking to a stop is still moving forward, but it’s accelerating backward (because the brake force is backward). Direction of motion ≠ direction of acceleration.

The Newton (the Unit)

One newton (N) is the force needed to accelerate a 1 kg mass at 1 m/s². So 1 N = 1 kg·m/s².

To get a feel for the size: an apple weighs about 1 N. A liter of water weighs about 10 N. A 70 kg adult weighs about 700 N. If you’re ever lost on whether your answer is “way too big” or “way too small,” anchor to these.

Weight vs. Mass

These two words are not interchangeable on the MCAT.

  • Mass (m) is how much stuff is in an object. Scalar. Measured in kilograms. Doesn’t change if you move to the moon, the bottom of the ocean, or the back of an ambulance.
  • Weight (W) is the gravitational force pulling that mass downward. Vector. Measured in newtons. Changes everywhere gravity changes.

Everyday version: when your bathroom scale says “150 lb,” that’s actually telling you the gravitational force on your body, not your mass. The MCAT holds you to the physics definitions strictly — if a question gives you “70 kg,” that’s mass; the corresponding weight is about 700 N.

Free-Body Diagrams

A free-body diagram (FBD) is a sketch showing every force acting on one specific object. It’s the most important problem-solving tool in dynamics — and skipping it is the #1 reason students miss force questions. Every MCAT force problem should start with a quick FBD.

How to draw one in 30 seconds:

  1. Draw the object as a box or dot. Ignore everything else in the scene.
  2. Draw an arrow for every force acting on the object, starting from the center.
  3. Label each arrow (weight, normal, friction, tension, push, etc.).
  4. Do NOT draw forces the object exerts on other things. Those go on the other object’s diagram.
Free-body diagram of a box on an inclined surface showing weight, normal force, applied force, and friction force vectors with labeled magnitudes and directions
Free-body diagram for a box on a surface. Each arrow represents one force acting on the box. Credit: Wikimedia Commons, CC BY-SA

Common forces on FBDs:

ForceSymbolDirectionNotes
WeightW or mgDownwardAlways present near Earth’s surface
Normal forceN or FNF_{N}Perpendicular to surfaceSurface pushes back on whatever’s touching it
FrictionfParallel to surface, opposing motionStatic or kinetic
TensionTAlong the rope/stringPulls, never pushes
Applied forceFappF_{\text{app}}Direction of push/pullGiven in the problem

Applying Newton’s Second Law

Once your FBD is drawn, the math is mechanical. Pick a coordinate system (usually x = horizontal, y = vertical), then add up the forces in each direction separately:

  • x-direction: sum of all x-forces = maxma_x
  • y-direction: sum of all y-forces = mayma_y

If the object isn’t accelerating in a direction (e.g., a box sitting on a table isn’t accelerating vertically — it’s just sitting there), then the net force in that direction is zero. That fact alone often unlocks the normal force or tension.

Worked example. You push a 30 kg suitcase across an airport floor with a 60 N horizontal force. Friction opposes the motion at 30 N. What’s the suitcase’s acceleration?

  • FBD has 4 forces: weight down, normal up (these cancel — no vertical motion), push (60 N forward), friction (30 N backward).
  • Net horizontal force = 60 − 30 = 30 N forward.
  • a=Fnet/m=30/30=1a = F_{net}/m = 30/30 = 1 m/s² forward.

That’s it. Three lines of work because we drew the diagram first.

Elevator Problems (Apparent Weight)

The MCAT loves elevator problems because they hide a really clean physics insight inside an everyday experience.

You’ve felt it: when an elevator starts going up, you feel briefly heavier. When it slows down at the top, you feel briefly lighter. Your actual weight (mg) hasn’t changed — but the floor (and the scale, if you’re standing on one) is pushing on you with a different force.

That force is the normal force, and the scale reads exactly that. We call it your apparent weight.

In an elevator accelerating upward, the floor has to do two jobs: hold you up and speed you up. So it pushes harder than usual. In an elevator accelerating downward, the floor still holds you up but doesn’t have to push as hard, because gravity is doing some of the work.

Elevator motionAccelerationNormal force (scale reading)You feel…
Accelerating upUpwardN = m(g + a) > mgHeavier
Constant velocity (up or down)0N = mgNormal
Accelerating downDownwardN = m(g - a) < mgLighter
Free fallg downwardN = 0Weightless

Notice that constant velocity feels normal — even at the top of a fast-moving elevator, if it’s cruising at steady speed, you feel just like standing on the ground. The “heavy” and “light” sensations only happen during the acceleration phases (start and stop).

A 60 kg person stands on a scale in an elevator accelerating upward at 2 m/s². What does the scale read? (Use g = 10 m/s²)
Click to reveal answer
720 N. The scale reads the normal force: N = m(g + a) = 60(10 + 2) = 60(12) = 720 N. The true weight is mg = 600 N, but the upward acceleration requires extra normal force, making the person feel heavier.
A 5 kg box is pushed with a 30 N horizontal force across a frictionless surface. What is its acceleration?
Click to reveal answer
6 m/s². Using Fnet=maF_{net} = ma: 30 N = 5 kg ×\times a, so a = 305\frac{30}{5} = 6 m/s² in the direction of the applied force. On a frictionless surface, the only horizontal force is the applied force, so it is also the net force.
An 80 kg person stands in an elevator that's slowing down as it rises. The deceleration is 1 m/s². What does the floor push up on them with? (g = 10)
Click to reveal answer
720 N. "Slowing down while rising" means the acceleration vector points downward (opposite to motion). So use N = m(g − a) = 80(10 − 1) = 720 N. The person feels lighter than their normal weight of 800 N — the same sensation you get when an elevator slows to a stop near the top floor.