Preview

Forces and Motion: Basics

Q: A 20kg box has μ_k = 0.3. What applied force is needed to accelerate it at 2 m/s²?

F_net = ma = 20×2=40N; F_applied = F_net + F_friction = 40 + 0.3×20×9.8.

Q: What is the minimum force to keep a 20kg box moving at constant velocity with μ_k = 0.3?

Constant velocity → F_net = 0 → F_applied = F_friction = μmg.

Q: Why does a heavier box accelerate the same as a lighter one at the same force/mass ratio?

F/m = a; if you scale both F and m by the same factor, a is unchanged.

Push objects and observe Newton's laws in action

For two thousand years, the smartest people alive believed that motion needs a motor. Aristotle said it plainly: keep pushing, or the cart stops. And every cart, sled, and shopping trolley you have ever pushed agrees with him — let go, and the thing grinds to a halt. It took Galileo's polished ramps and Newton's first law to catch the impostor: the cart doesn't stop because motion runs out; it stops because an unseen force — friction — actively kills it. Zero net force doesn't mean zero motion, it means zero CHANGE in motion. A hockey puck on smooth ice glides on; Voyager 1 has been coasting engine-off since 1980 and hasn't slowed a step. In this lab the fraud is exposed live: push a block down a frictionless track, then yank the force away mid-glide — the acceleration row snaps to zero and the velocity row simply freezes, the block cruising and bouncing between the bumpers with nothing pushing it at all. You'll also catch static friction red-handed matching your push newton-for-newton up to its threshold, watch a rough slope bill mechanical energy into a climbing Heat bar, and run an Atwood machine where two hanging masses negotiate one rope tension. By the end, F = ma will read the way Newton meant it: force is the price of changing motion — never the rent for keeping it.

What you'll be able to do

Split Newton's first and second laws with the panel: at the flat default (F=15 N, m=2 kg, µ=0.2) read f=3.9 N, F_net=11.1 N, a=5.54 m/s²; then on the frictionless Heavy vs Light state drag the push to zero mid-glide and read a=0.00 with v frozen — no force is needed to keep moving, only to change the moving
Read static friction as a responsive force on the f row: below threshold (F=3 N < µmg=3.92 N) friction matches the push exactly — f=3.0 N, a=0.00 — and only at breakaway does it settle to the kinetic µN=3.9 N (fs ≤ µsN vs fk = µkN, the sim's two friction regimes on one row)
Carry F_net = ma across scenarios: the frictionless 20° incline runs a=−3.35 m/s² with N=18.4 N and zero heat (KE↔PE conserved); the rough 25° incline bleeds mechanical energy into the climbing Heat bar; the 1 kg/3 kg Atwood machine accelerates at (m₂−m₁)g/(m₁+m₂)=4.90 m/s² with rope tension 14.7 N

Formulas

F_{net} = ma

Newton's Second Law

F_{friction} = \mu_k mg

Kinetic friction force

F_{net} = F_{applied} - F_{friction}

Net force

a = g(\sin\theta - \mu_k\cos\theta)

Acceleration down a slope with friction

N = mg\cos\theta

Normal force on an incline of angle θ

a = \frac{(m_2 - m_1)g}{m_1 + m_2}

Atwood machine acceleration (two hanging masses)

Make a prediction

Heavy vs Light preset: a 10 kg block on the frictionless flat track (µ = 0), pushed with 50 N — the a row reads 5.00 m/s² and the block is gliding along at several m/s. Now drag Applied Force down to 0 while it moves. What does the block do?

No grading here — commit to a guess, then scroll down and test it yourself.

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Explore

Run the gate live: Heavy vs Light preset, let the block wind up to speed under the 50 N push, then drag Applied Force to 0 mid-glide. Watch three rows at once — a snaps to 0.00, F_net to 0.0, and v refuses to change — while the block coasts and bounces, unpowered. That frozen v row is Newton's first law on a screen.
Catch static friction adjusting: back on Push on Flat, walk the force up from 0 → 3 → 5 N and watch the f row answer 0.0 → 3.0 → 3.9. Below the 3.92 N threshold friction matches your push exactly (a stays 0.00); at 5 N the block breaks loose and friction settles to its kinetic µN. One row, two regimes — fs ≤ µN is an inequality, not an equation.
Race the two inclines: Frictionless Incline trades KE and PE back and forth with the Heat bar dead at 0.0 J, forever; Rough Incline (µ=0.3) slides down at −1.48 m/s² while the Heat bar climbs and the Mechanical bar shrinks by exactly that much. Same hill, one bookkeeping difference — friction is a tax collector, not a thief.
Interrogate the Atwood machine: at 1 kg vs 3 kg it runs a=4.90 m/s² with T=14.7 N — strictly between the two weights (9.8 and 29.4 N), because the rope must slow the heavy side and drag the light side. Now slide the masses equal and watch a=0.00 with T=mg: a balanced Atwood hangs still, each side carrying its own weight.

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Check

Did the lab agree with you?

Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: It keeps gliding at whatever velocity it had — a snaps to 0.00, v freezes, and it coasts between the bumpers forever; with zero net force, motion simply continues (Newton's first law)

With the push at zero and µ = 0, the net force on the block is exactly zero — and zero net force means zero acceleration, not zero motion: 'A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless acted on by a net external force' (OpenStax §5.2, verbatim). On the panel: a snaps to 0.00, F_net to 0.0, and v holds its value indefinitely — the block coasts, bouncing elastically off the bumpers, unpowered. Options B and C are the two classic strengths of the same documented preconception — 'motion implies a force' (Clement 1982), the belief the FCI's first-law cluster was built to catch. It feels true because every real push you've ever quit was followed by friction quietly doing the stopping: 'Friction is thus the cause of slowing... The object would not slow down if friction were eliminated.' This sim eliminates it, and the block calls the 2000-year bluff. Force is the price of CHANGING motion — never the rent for keeping it.

Quiz (0/3)

A 20kg box has μ_k = 0.3. What applied force is needed to accelerate it at 2 m/s²?

What is the minimum force to keep a 20kg box moving at constant velocity with μ_k = 0.3?

Why does a heavier box accelerate the same as a lighter one at the same force/mass ratio?

You can now

Split Newton's first and second laws with the panel: at the flat default (F=15 N, m=2 kg, µ=0.2) read f=3.9 N, F_net=11.1 N, a=5.54 m/s²; then on the frictionless Heavy vs Light state drag the push to zero mid-glide and read a=0.00 with v frozen — no force is needed to keep moving, only to change the moving
Read static friction as a responsive force on the f row: below threshold (F=3 N < µmg=3.92 N) friction matches the push exactly — f=3.0 N, a=0.00 — and only at breakaway does it settle to the kinetic µN=3.9 N (fs ≤ µsN vs fk = µkN, the sim's two friction regimes on one row)
Carry F_net = ma across scenarios: the frictionless 20° incline runs a=−3.35 m/s² with N=18.4 N and zero heat (KE↔PE conserved); the rough 25° incline bleeds mechanical energy into the climbing Heat bar; the 1 kg/3 kg Atwood machine accelerates at (m₂−m₁)g/(m₁+m₂)=4.90 m/s² with rope tension 14.7 N