Newton's Second Law (F=ma) Simulator

push a block, watch F = ma play out — friction, an incline, and a full free-body diagram

Starting values — Newton's Second Law (F=ma) Simulator

Mass (m)	2
Applied force (F)	10
Friction coefficient (μ)	0.2
Incline angle (θ)	0
Initial velocity (u)	0

Change any control above and the results recompute from these.

Loading interactive simulation...

why weight and normal force aren't a 3rd-law pair 🖖

Weight (mg) and the normal force (N) are equal and opposite here, so it's tempting to call them a Newton's-third-law pair — but they aren't. A third-law pair must act on two *different* objects, and both of these act on the same block. The real pairs are: Earth pulls the block down / the block pulls Earth up (gravity, both directions), and the block pushes down on the surface / the surface pushes back up on the block — that second one is the N you see here. N only happens to equal mg because the block isn't accelerating vertically; tilt the surface or put the block in an accelerating elevator and N changes while mg doesn't. This exact mix-up is one of the most well-documented misconceptions in introductory mechanics.

Only the leftover force accelerates 🖖

Newton's second law, F=ma, is really about the net force — everything else cancels out. When you push this block, part of your push can be eaten by friction; only what's left over drives the acceleration. That's why a 10 N push on a 2 kg block gives a = 5 m/s² with no friction, but less once friction takes its cut. Double the mass and the same net force gives half the acceleration.

Newton never actually wrote F=ma 🖖

The equation stamped on this tool isn't quite Newton's. In the Principia (1687) he stated the second law in words, in terms of momentum: the change in a body's quantity of motion is proportional to the impressed force. The tidy algebraic form F = ma came later — it's usually credited to Leonhard Euler around 1752. For constant mass the two agree, but it's the momentum version (F = dp/dt) that survives into relativity and rockets shedding fuel as they burn.

F = ma WITH FRICTION — WILL IT MOVE, AND WHICH FRICTION APPLIES?

Which Force Case Are You Solving?

Almost every mistake in this topic happens before any arithmetic: people write down f = μN for a block that is not moving, or look for a force to explain motion that needs none. Two comparisons settle every case here. Is the driving force bigger than the friction ceiling μN — and is the block already moving? Answer those two and the acceleration follows in one line.

Clean second law — a push and nothing else a = F/m = 5 m/s²

Nothing happens — the push loses to static friction |F| ≤ μN ⇒ a = 0

It moves, and friction keeps taking its cut a = (F − μN)/m = 1.04

Moving with zero net force — the first law ∑F = 0 ⇒ a = 0, v = 6

Friction alone — how far before it stops a = −μg = −2.94 m/s²

On a slope — tan θ against μ tan θ > μ ⇒ a = g(sin θ − μ cos θ)

Clean second law — a push and nothing else

What you know: Flat ground, no friction at all (μ = 0), block starting from rest. The applied force is the only horizontal force on it.

How it resolves: a = F/m = 5 m/s²

Worked example: m = 2 kg, F = 10 N → the friction ceiling is 0 N, so the net force is the full 10 N and a = 10/2 = 5 m/s². Over the tool's 6 s window the block reaches 30 m/s and travels 90 m.

Open this case: frictionless push

One horizontal arrow, no friction: the net force is F itself and a = F/m exactly.

Nothing happens — the push loses to static friction

What you know: Flat ground, block at rest, a rough surface (μ = 0.6) and a modest push. Before anything else, compare the push with the ceiling μN.

How it resolves: |F| ≤ μN ⇒ a = 0

Worked example: m = 10 kg, F = 5 N, μ = 0.6 → N = 98.1 N so the ceiling is μN = 58.86 N. The 5 N push is nowhere near it, so friction answers with exactly −5 N, the net force is 0, a = 0, and the block is still at x = 0 after 6 s.

Open this case: stuck by friction

Nothing happens — the push loses to static friction. Friction matches the push exactly — 5 N, not the 58.86 N ceiling. Flat ground, block at rest, a rough surface (μ = 0.6) and a modest push. Before anything else, compare the push with the ceiling μN. — Friction matches the push exactly — 5 N, not the 58.86 N ceiling.

It moves, and friction keeps taking its cut

What you know: Flat ground again, but now the push clears the ceiling. Once the block is sliding, friction stops adjusting: it sits at μN and points against the motion.

How it resolves: a = (F − μN)/m = 1.04

Worked example: m = 5 kg, F = 15 N, μ = 0.2 → N = 49.05 N, ceiling 9.81 N. Since 15 N > 9.81 N the block breaks free, friction locks at 9.81 N, the net force is 15 − 9.81 = 5.19 N and a = 1.038 m/s². After 6 s: 6.23 m/s, 18.68 m.

Open this case: push with friction

It moves, and friction keeps taking its cut. Friction is now fixed at μN = 9.81 N; the leftover 5.19 N is what accelerates. Flat ground again, but now the push clears the ceiling. Once the block is sliding, friction stops adjusting: it sits at μN and points against the motion. — Friction is now fixed at μN = 9.81 N; the leftover 5.19 N is what accelerates.

Moving with zero net force — the first law

What you know: No applied force, no friction, but the block is already travelling at 6 m/s when the clock starts.

How it resolves: ∑F = 0 ⇒ a = 0, v = 6

Worked example: m = 2 kg, u = 6 m/s, F = 0, μ = 0 → every force row reads zero, the net force is 0 and a = 0. Six seconds later it is still doing 6 m/s, now 36 m further along.

Open this case: no net force

Every force is zero and the block keeps its 6 m/s — for exactly that reason.

Friction alone — how far before it stops

What you know: Nobody is pushing. The block is sliding at 8 m/s on a surface with μ = 0.3, and friction is the only horizontal force.

How it resolves: a = −μg = −2.94 m/s²

Worked example: m = 3 kg, u = 8 m/s, μ = 0.3 → friction 8.83 N, a = −2.94 m/s². It comes to rest at t = u/(μg) = 2.72 s having covered u²/(2μg) = 10.87 m, and then stays there.

Open this case: sliding to a stop

Friction alone — how far before it stops. 8 m/s gone in 2.72 s and 10.87 m — and the mass never enters the answer. Nobody is pushing. The block is sliding at 8 m/s on a surface with μ = 0.3, and friction is the only horizontal force. — 8 m/s gone in 2.72 s and 10.87 m — and the mass never enters the answer.

On a slope — tan θ against μ

What you know: A block released from rest on a 20° incline with μ = 0.1 and nobody pushing. Gravity now has a component along the surface, and the normal force is no longer the full weight.

How it resolves: tan θ > μ ⇒ a = g(sin θ − μ cos θ)

Worked example: m = 4 kg, θ = 20°, μ = 0.1 → N = mg cos θ = 36.87 N (not the 39.24 N weight), gravity along the slope = 13.42 N, ceiling μN = 3.69 N. Since 13.42 N > 3.69 N it slides: a = g(sin θ − μ cos θ) = 2.43 m/s², reaching 43.8 m down the slope in 6 s.

Open this case: slides down a ramp

On a slope — tan θ against μ. Gravity along the slope (13.42 N) against a ceiling of 3.69 N — it slides. A block released from rest on a 20° incline with μ = 0.1 and nobody pushing. Gravity now has a component along the surface, and the normal force is no longer the full weight. — Gravity along the slope (13.42 N) against a ceiling of 3.69 N — it slides.

References (2)

Insight block 1 — the weight/normal-force pairing is not a casual mistake but a measured one: D. Hestenes, M. Wells and G. Swackhamer, "Force concept inventory." The Physics Teacher 30(3), 141–158, 1992 — the instrument that established how widespread these particular confusions are.
Insight block 3 — the second law as Newton actually stated it: I. Newton, Philosophiæ Naturalis Principia Mathematica, Book I, Law II. London, 1687.

Problem solved in full

A 2 kg block on level ground pushed with 10 N 5 steps
A 2 kg block on level ground, pushed with 10 N, on a surface with a friction coefficient of 0.2. Find whether it moves at all, then how far it travels in 6 s — and then find the mass that would defeat the same push.
1. Friction is set by how hard the surfaces are pressed together, not by how hard you shove them along. On level ground that press is the block's whole weight.
  
  $N = m g cos θ = 2 \times 9.81 \times cos 0^{\circ} = 19.62 N$
2. Static friction is not a fixed force. It supplies whatever is asked of it up to a ceiling of μN, and here the ceiling is well under the applied 10 N, so the block slips. Settle this comparison before computing any acceleration, because stuck and sliding obey different equations.
  
  $f_{s, m a x} = μ N = 0.2 \times 19.62 = 3.924 N < F = 10 N$
3. Once it is sliding, friction stops adjusting and takes its full value against the motion. Only the leftover accelerates anything.
  
  $F_{net} = F - μ N = 10 - 3.924 = 6.076 N$
4. Now the second law does its one job, and a constant acceleration feeds straight into the displacement. The block starts from rest, so all 54.684 m of it comes from the acceleration term.
  
  $a = \frac{F _{net}}{m} = \frac{6.076}{2} = 3.038 m/ s^{2}, x = \frac{1}{2} a t^{2} = \frac{1}{2} (3.038) (6)^{2} = 54.684 m$
5. Mass enters the answer twice and with opposite signs: it divides the push and it multiplies the friction. The acceleration therefore falls faster than 1/m, and unlike 1/m it reaches zero at a finite mass.
  
  $a (m) = \frac{F}{m} - μg = \frac{10}{m} - 1.962 : a (4 kg) = 0.538 m/ s^{2}, a = 0 ⟺ m = 5.097 kg$
Answer

The tool prints a net force of 6.076 N, an acceleration of 3.038 m/s² and 54.684 m covered in 6 s. Double the mass to 4 kg and the acceleration does not halve: it drops to 0.538 m/s², a factor of 5.6, because friction has grown from 3.924 N to 7.848 N while the push stayed at 10 N. At 5.097 kg the two are equal and nothing moves, so every push has a mass ceiling of F/μg. That is why heavy things get a trolley or a film of grease rather than a stronger pusher — μ and F sit on opposite sides of the same fraction, so halving the friction coefficient lifts that ceiling exactly as much as doubling the force does.

$a = \frac{F}{m} - μg cos θ (θ = 0)$

By the 42Math team · Updated 2026-08-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

no net force - No net force on a moving block → constant velocity forever (Newton's 1st law)
frictionless push - m=2 kg, F=10 N, no friction → a=5 m/s² (clean 2nd law)
push with friction - m=5 kg, F=15 N, mu=0.2 → friction cuts the net force, a≈1.04 m/s²
stuck by friction - Max static friction (58.9 N) beats the 5 N push → block stays put, a=0
sliding to a stop - Friction alone brings a moving block to rest in ~2.7 s, then it stays stopped
slides down a ramp - 20° incline: gravity (13.4 N) beats max static friction (3.7 N) → slides from rest