Pick your class, read the steps, then try them right in the lab —
change one slider at a time and watch the arrows and graphs respond.
Class 9 — Newton's first law: inertia
Concept: a body keeps its state of rest or of steady motion in a straight line unless a net
force acts on it. This tendency to keep going is called inertia.
- Open Force & motion. Set Applied force = 0, Friction μ = 0 and Starting velocity = 4 m/s.
Play: the block carries on at 4 m/s for ever. The forces (weight and normal force) cancel, so the net force is zero.
- Now raise Friction μ to 0.2. Friction is the only horizontal force, so the block slows down:
a = μg = 0.2 × 9.8 = 1.96 m/s². It stops after 4 ÷ 1.96 = 2.04 s, having travelled 4.08 m.
- Set Starting velocity back to 0 and Applied force = 4 N with μ = 0.2 (the greatest friction is
0.2 × 5 × 9.8 = 9.8 N): the push is smaller, so friction just balances it and the block stays at rest.
Takeaway: motion does not need a force to keep it going; it is a change of motion that
needs one. Friction is the hidden force that makes things stop.
Class 9 — Newton's second law: F = ma
Concept: the net force on a body equals its mass times its acceleration, ΣF = ma. A bigger
force gives a bigger acceleration; a bigger mass gives a smaller one.
- In Force & motion set Force = 30 N, Mass = 5 kg, μ = 0, Starting velocity = 0.
a = 30 ÷ 5 = 6.0 m/s²; after 5 s the speed is 30 m/s.
- Double the mass to 10 kg: a = 3.0 m/s². The same force on twice the mass gives half the acceleration.
- Back to 5 kg and add friction, μ = 0.2. The friction is 0.2 × 5 × 9.8 = 9.8 N, so the
net force is 30 − 9.8 = 20.2 N and a = 20.2 ÷ 5 = 4.04 m/s².
- Look at the graphs: the speed rises on a straight line (constant acceleration) and the acceleration graph is flat.
Takeaway: use the net force: add up all the forces along the motion (the push, minus
friction) and divide by the mass.
Class 11 — Static and kinetic friction
Concept: friction on a body at rest (static) adjusts itself to match the push, up to a limit
μsN. Once the body slides, the friction is kinetic, μkN, and it is smaller.
- Open Static & kinetic friction, choose Wood on wood (μs = 0.5, μk = 0.3),
mass 5 kg, pull growing at 10 N/s. Here N = mg = 49 N.
- The friction rises along the straight line of the graph, always equal to the pull. The block does not move
until the pull reaches μsN = 0.5 × 49 = 24.5 N, after 2.45 s.
- The instant it moves the friction drops to μkN = 0.3 × 49 = 14.7 N. The pull is
still 24.5 N, so a = (24.5 − 14.7) ÷ 5 = 1.96 m/s².
- Try Ice on steel: the break-away force is only 0.03 × 49 = 1.47 N. Try Rubber on dry concrete: 49 N.
- Change the pull rate: the block still breaks free at the same force, only at a different time. The area of contact
does not appear in any formula.
Takeaway: it is harder to start a heavy object moving than to keep it moving, because
μs > μk.
Class 11 — A block on a rough slope and the angle of repose
Concept: split the weight into mg sinθ down the slope and mg cosθ into it. Friction up to
μsmg cosθ can hold the block, so it slides only when tanθ > μs.
- Open Block on an incline with mass 5 kg, μs = 0.5, μk = 0.4 and no push.
Tick the weight components to see them.
- At θ = 25° the block stays put: it needs only mg sin 25° = 20.71 N of friction, less than the
greatest 0.5 × 42.4 = 21.2 N.
- The angle of repose is tan−1 μs = tan−1 0.5 = 26.6°. On the left graph this is where the
orange curve (friction needed) crosses the red one (friction available).
- At θ = 30° the block slides with a = g(sin 30° − 0.4 cos 30°) = 1.51 m/s², and it
reaches the bottom of the 8 m slope after about 3.3 s.
- Change the mass: the angle at which it starts to slide does not change, because the mass cancels.
Takeaway: measuring the angle at which a block just starts to slip gives μs = tanθ.
Class 11 — How much push on a slope?
Concept: a push P up the slope reduces the force needed from friction, and a bigger push can drive
the block up the slope, where friction then points down.
- In Block on an incline set θ = 30°, mass 5 kg, μs = 0.5, μk = 0.4.
With no push the block slides (30° is steeper than the 26.6° angle of repose).
- To hold it you need at least P = mg(sin θ − μs cos θ) =
49 × (0.5 − 0.433) = 3.28 N up the slope. Try 3 N (it slides) and 4 N (it stays).
- To make it move up you need more than mg(sin θ + μs cos θ) =
49 × (0.5 + 0.433) = 45.72 N, because friction now acts down the slope too.
- Between the two values the block does not move at all: static friction can point either way.
Takeaway: static friction is flexible: it can take any value up to its limit, in whichever direction stops the motion.
Class 11 — Connected bodies, tension and contact forces
Concept: bodies joined by a taut string, or pushing each other, share one acceleration. Find it from the
whole system, then find the internal force (tension or contact force) from one body alone.
- Open Strings, contact forces & lifts, first situation: A = 4 kg on the table, B = 2 kg hanging, μ = 0.2.
a = (mBg − μmAg) ÷ (mA + mB) = (19.6 − 7.84) ÷ 6 = 1.96 m/s².
- The tension is T = mB(g − a) = 2 × (9.8 − 1.96) = 15.68 N: less than B's weight (19.6 N)
because B accelerates downwards. Check with A: μmAg + mAa = 7.84 + 7.84 = 15.68 N.
- Switch to Two blocks pushed in contact: m1 = 3 kg, m2 = 2 kg, F = 40 N, μ = 0.1.
a = (40 − 4.9) ÷ 5 = 7.02 m/s².
- Block 2 alone: N = m2(a + μg) = 2 × (7.02 + 0.98) = 16 N. Block 1 pushes block 2 with 16 N and,
by Newton's third law, block 2 pushes block 1 back with exactly 16 N. The pair acts on different bodies, so
they do not cancel.
Takeaway: action and reaction are equal and opposite but act on different bodies; the internal
force is smaller than the applied one because it only has to move part of the load.
Class 11 — Apparent weight in a lift
Concept: a scale reads the normal force N, not the true weight. For a person of mass m in a lift
accelerating upwards at a, N − mg = ma, so N = m(g + a).
- In Strings, contact forces & lifts choose the lift, person 60 kg, a = 2 m/s². The true weight is
mg = 588 N.
- While the lift speeds up, the scale reads 60 × (9.8 + 2) = 708 N: you feel 20% heavier. At the steady
speed it reads 588 N. While it slows down the reading is 60 × (9.8 − 2) = 468 N: you feel lighter.
Watch the needle and the lower graph.
- Going down is the same with the two end phases swapped, because only the direction of the acceleration
matters, not of the velocity.
- Set a = −9.8 m/s² (the cable snaps): N = m(g − g) = 0. The scale reads zero, the person is
weightless, even though gravity has not gone away: this is the state of astronauts in orbit.
Takeaway: the weight you feel is the push of the floor on you, N = m(g + a).