Pick your class, read the steps, then try them right in the lab —
change one slider at a time and watch the energy bars and graphs respond.
Class 9 — Work: force × displacement
Concept: when a force moves something in the direction of the force, it does
work: W = F × d, measured in joules (1 J = 1 N·m). No movement, no work.
- Open Work by a force. Set Force F = 20 N, Angle = 0°, Distance = 5 m, Block mass = 5 kg,
Friction = 0 and Starting speed = 0. Play.
- The lab shows W = 20 × 5 = 100 J. The block ends with 100 J of kinetic energy, so
its speed is √(2 × 100 ÷ 5) = 6.32 m/s: the work done is the energy it gained.
- Double the force to 40 N: the work doubles to 200 J. Double the distance instead (10 m): again
double the work. Work is proportional to both.
- Look at the left graph: the shaded rectangle (force × distance) is the work.
Takeaway: work = force × distance moved in the direction of the force, and it is
the energy transferred to the object.
Class 11 — Work at an angle: positive, zero and negative
Concept: only the part of a force along the motion does work:
W = F d cosθ. A force at right angles to the motion does none, and a force against the
motion does negative work (it removes energy).
- In Work by a force set F = 40 N, d = 5 m, m = 5 kg, Friction = 0, Starting speed = 0,
and the angle to 0°. W = 40 × 5 × cos 0° = 200 J.
- Set the angle to 60°: only F cos 60° = 20 N works along the floor, so
W = 40 × 5 × cos 60° = 100 J. The block now reaches 6.32 m/s, not 8.94 m/s.
- Set the angle to 90°: cos 90° = 0, so W = 0. The block does not even start to
move, and the worked panel says no work is done.
- Set the starting speed to 6 m/s and the angle to 180° (the force pushes against the
motion). The work is negative; the block loses its 90 J of kinetic energy and stops after
s = v² ÷ 2|a| = 36 ÷ 16 = 2.25 m. Check: Wnet = −90 J = ΔKE.
Takeaway: work is a scalar that can be positive, zero or negative, depending on the angle
between the force and the displacement.
Class 9 — Kinetic and potential energy turn into each other
Concept: an object at a height has potential energy PE = mgh. As it falls the PE
becomes kinetic energy KE = ½mv²; with no friction the sum never changes.
- Open Work–energy on a slope and set Friction μ = 0, height h = 3 m, mass = 2 kg.
Play and watch the bars: PE falls, KE rises, the total stays the same.
- At the bottom all of mgh = 2 × 9.8 × 3 = 58.8 J has become kinetic energy:
½mv² = mgh, so v = √(2gh) = 7.67 m/s.
- Change the slope angle from 20° to 60°. The speed at the bottom is still 7.67 m/s:
it depends only on the height, not on the shape of the path.
- Change the mass to 10 kg. The speed is the same again — mass cancels.
Takeaway: without friction, PE + KE is constant, and a falling object's speed depends only
on the height it has fallen through.
Class 11 — The work–energy theorem and friction
Concept: the total work done by all the forces equals the change in kinetic energy,
Wnet = ΔKE. Friction does negative work and turns mechanical energy into heat.
- In Work–energy on a slope set α = 30°, h = 3 m, μ = 0.2, m = 2 kg.
The slope is L = 3 ÷ sin 30° = 6 m long.
- Gravity does mgh = 58.8 J of work; friction does −μmg cosα × L = −20.37 J.
Their sum is the kinetic energy at the bottom, 38.43 J, so v = 6.2 m/s (less than the 7.67 m/s of
the smooth slope).
- On the floor only friction acts: −μmg × D = 0 − 38.43 J gives
D = 9.8 m. Play and watch the block stop there.
- Add up the heat: 20.37 J on the slope + 38.43 J on the floor = 58.8 J, exactly the starting
potential energy. The dashed total on the energy graph never changes.
- Raise μ to 0.6: tan 30° = 0.577 is less than 0.6, so the block does not slide at all.
Takeaway: with friction the mechanical energy is not conserved, but the total energy
(including heat) is. A block slides only if tanα > μ.
Class 11 — Energy conservation and the loop-the-loop
Concept: on a frictionless track KE + PE is constant. At the top of a vertical loop the
track can only push, never pull, so the ball needs mv²÷R ≥ mg there, i.e. v² ≥ gR.
- Open Energy conservation & the loop with R = 1.5 m and h = 6 m. The ball completes
the loop. At the bottom v = √(2gh) = 10.84 m/s.
- At the top (height 2R = 3 m) energy conservation gives
v = √(2g(h − 2R)) = √(2 × 9.8 × 3) = 7.67 m/s. The track pushes with
N = mv²÷R − mg = 3mg = 29.4 N for a 1 kg ball.
- The least height that works: put v² = gR at the top, 2g(h − 2R) = gR, so
h = 2.5R = 3.75 m. Try h = 3.8 m (it just makes it) and h = 3.7 m (it leaves the track near the top).
- Try h = 3 m: between R and 2.5R. The contact force reaches zero at
cosφ = (2R − 2h) ÷ 3R = −0.667 (φ = 131.8° from the bottom) and the ball
flies off as a projectile.
- Try h = 1.2 m: not above the centre of the loop, so the ball just slides back and forth.
Change the mass: nothing changes, because it cancels.
Takeaway: for a ball sliding on a vertical loop the minimum release height is 2.5R, and the
minimum speed at the bottom is √(5gR).
Class 9 — Power: how fast work is done
Concept: power is the rate of doing work, P = W ÷ t. The unit is the watt
(1 W = 1 J/s). A bigger power does the same work in less time.
- Open Power and choose A motor lifting a load: M = 100 kg, P = 2 kW,
efficiency 80%, height 10 m. Play.
- The work needed to lift the load is Mgh = 100 × 9.8 × 10 = 9800 J. The useful power
is 0.8 × 2000 = 1600 W, so the load rises at v = 1600 ÷ (100 × 9.8) = 1.63 m/s
and takes 6.13 s.
- The motor draws 2000 W for 6.13 s = 12.25 kJ = 0.0034 kWh from the supply, but only 9.8 kJ ends up
as potential energy; the rest is lost as heat.
- Raise the efficiency to 100%: the time drops to 4.9 s and nothing is wasted.
Takeaway: power = work ÷ time; efficiency = useful output ÷ input; and
1 kWh = 3.6 million joules.
Class 11 — P = Fv: the power of a car and its top speed
Concept: for a force along the motion, P = F v. At a constant engine power the pull
F = P ÷ v falls as the car speeds up, and the car stops accelerating when F equals the resistance.
- In Power choose A car accelerating on a road (60 kW, 1200 kg). At 10 m/s the engine
pulls with F = 60000 ÷ 10 = 6000 N.
- The resistance is 176 N of rolling resistance plus 0.42v² of air drag. The top speed solves
60000 ÷ v = 176 + 0.42v², giving 49.6 m/s (179 km/h). On the force–speed graph it is
where the two curves cross.
- Double the power to 120 kW. The top speed does not double: it rises only to 63.7 m/s (229 km/h),
because drag grows as v².
- Return to 60 kW and set the slope to 5°. The weight now adds 1025 N of resistance, so the top
speed falls to 35 m/s (126 km/h).
Takeaway: P = F v; at top speed the engine force equals the resistance, so
vtop = P ÷ Fresist when the resistance is constant.