Track mechanical energy during a descent

Choose the system before balancing the energy

Consider the particle together with its gravitational interaction with Earth. We represent that interaction using gravitational potential energy U. The particle’s translational motion contributes kinetic energy K. In this ideal model, a fixed frictionless ramp guides the particle but does no work: its normal force is perpendicular to each small displacement along the ramp.

There is no air drag, friction, motor, moving support or rotation. Under these assumptions, the mechanical energy E = K + U remains constant. This is conservation during one physical descent, not a claim that the number E stays the same when we change the accounting reference between separate descriptions.

Use reference height zero first. The 2 kg particle starts from rest at height 4 m. With teaching g=10 m/s², its initial potential is Uinitial = 2 × 10 × 4 = 80 J and initial kinetic energy is zero. Total mechanical energy is therefore 80 J.

Follow the exchange through three heights

At height 2 m, potential energy is U = 2 × 10 × 2 = 40 J. Because the total remains 80 J, kinetic energy must be K = 80 − 40 = 40 J. The two middle bars in the chart show these parts; the total bar is their sum, not an additional energy store to add again.

At height zero, just before any interaction with the ramp end, potential is zero and kinetic energy is 80 J. The model stops its prediction at that state. It does not simulate hitting a floor, leaving the ramp, changing direction at a corner or coming to rest after an impact.

The speed follows from K = ½mv². At height 2 m, v = √(2 × 40 / 2) = √40 ≈ 6.325 m/s. At height zero, v = √80 ≈ 8.944 m/s. The square root matters: doubling kinetic energy does not double speed. With fixed mass, it multiplies speed by √2.

Capture the initial height, halfway height and final height. Explain the energy exchange before relying on a formula: as height falls, gravitational potential decreases while kinetic energy increases by the same amount. The total does not require an ongoing supply of energy from the normal force.

Reconcile two valid accounting methods

If the particle alone is the selected object, gravity does external work on it. The work-energy theorem gives ΔK = Wgravity, since the ramp normal does zero work. For a descent from 4 m to 2 m, gravity work is mg(4 − 2) = 40 J.

Alternatively, include the particle–Earth interaction in potential energy. Then Wgravity = −ΔU, and the same statement becomes ΔK + ΔU = 0, or Kinitial + Uinitial = Kfinal + Ufinal.

These are two descriptions of the same transfer. Do not decrease U by 40 J and then add another 40 J of gravity work to the right-hand side of that same conservation equation. Doing so counts gravity’s effect twice and predicts too much kinetic energy. State your system and keep the work or potential representation consistent.

The OpenStax conservation-of-energy treatment provides background for this accounting. The widget’s scenarios are original synthetic examples, not measured ramp trials.

Separate the speed prediction from the route and travel time

For a descent from rest with gravity as the only force doing work, energy gives ½mv² = mg(hinitial − h). Positive mass cancels, leaving v = √(2gΔh), where Δh is the positive vertical drop. The same drop produces the same speed under these assumptions, independent of mass.

The selected ramp is straight and tilted 30° below horizontal. Its down-slope acceleration is g sin 30° = 5 m/s². Distance along it is s = Δh/sin 30° = 2Δh. Starting from rest, time follows from s = ½at²: t = √(2s/a).

For a 2 m vertical drop, s = 4 m and t = √(8/5), approximately 1.265 s. Multiplying a × t gives approximately 6.325 m/s, agreeing with the energy prediction. The ramp geometry supplies the time calculation; energy alone did not determine that time.

A different frictionless route between the same heights can give the same endpoint speed but a different travel time, provided the particle actually follows that route and reaches the endpoint. Energy accounting alone does not establish contact everywhere, overcome an intermediate height barrier, or determine a complete trajectory. This widget uses a descending straight ramp to keep those conditions explicit.

Know when mechanical energy needs more terms

A rough ramp can convert some mechanical energy into internal energy of the contacting materials. A rolling object can store rotational kinetic energy as well as translational kinetic energy. A moving support or a motor can transfer energy into or out of the selected system. In such cases, the simple K+U balance used here is incomplete.

Total energy conservation is broader than conservation of mechanical energy. If K+U decreases, identify energy transferred across the system boundary or converted into other forms rather than declaring that energy vanished. The current model has no friction or thermal-energy control; those situations require an expanded accounting model.

Transfer: A 1 kg particle starts from rest at height 3 m. At height 1 m, an independent measurement reports that 5 J has been converted from mechanical energy into internal energy of the particle and rough ramp. With g=10 m/s², find the particle’s kinetic energy and speed. State the assumptions still needed.

Check the mechanical-energy transfer

Gravitational potential decreases by 1 × 10 × (3−1) = 20 J. With 5 J converted into internal energy and no other transfers or energy stores, translational kinetic energy is 15 J. Speed is √(2 × 15 / 1) = √30 ≈ 5.477 m/s. This assumes the particle reaches the endpoint, has no rotational kinetic energy, and the reported 5 J covers all nonmechanical conversion relevant to the balance.