Work, Energy & Power Simulator — Hoist Lift Energy Accounting Interactive

Interactive 3D hoist laboratory lifting a suspended load through a controlled rest-to-rest trajectory, tracking cable tension, gravitational and kinetic energy, mechanical power and electrical input through a drive of adjustable efficiency, with guided experiments, a model-verification bench and a knowledge-check quiz.

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About the Work, Energy & Power Simulator

This simulator lifts a suspended mass through a smooth rest-to-rest vertical motion using a controlled hoist drive of adjustable efficiency, tracking cable tension, gravitational and kinetic energy, mechanical and electrical power, and drive losses at every instant from launch to a complete stop.

What the simulator shows

• A real-time 3D gantry with guide rails, a motor/gearbox/drum drive, a cable and fixed sheave, a guided lifting cradle and load, and an electrical/mechanical energy display comparing cumulative input, work and loss — with home view, focus-selected-part, auto-rotate, expand and toggleable labels. • Four live controls: suspended mass (1–50 kg), lift height (0.5–5 m), lift duration (3–15 s) and drive efficiency (0.5–1, mechanical output divided by electrical input). • Eleven live metrics: height, lift speed, vertical acceleration, cable tension, gravitational energy gained, kinetic energy, mechanical work, mechanical power, electrical input power, electrical energy and drive energy loss. • A Curves & measurements tab charting energy accounting and power through the lift, the full model equations and snapshot measurements. • An Experiments tab with four guided scenarios (baseline lift, faster lift, doubling the load, ideal drive), a model-verification bench of independent automated checks, and a timestamped event log with a copyable trial report. • A Learn & assess tab with guided lessons, a two-question knowledge-check quiz and a written scope/reference statement.

Why mgh alone only works for the completed lift

Cable work is the integral of tension with respect to height, and tension itself changes throughout the smooth rest-to-rest trajectory as the load accelerates and decelerates: T = m(g + a). Mid-lift, using mgh alone would miss the kinetic energy the load has picked up; the correct running total is mechanical work = gravitational potential energy gained + kinetic energy. Only once the load returns to rest at the top — kinetic energy back to zero — does mechanical work settle to exactly mgh, which the baseline experiment confirms at 294.3 J for a 10 kg load lifted 3 m in 6 s at 80% efficiency.

Same work, different power — and where the electrical input goes

Instantaneous mechanical power is tension times upward velocity, so compressing the same 3 m lift from 6 s down to 3 s leaves the final mechanical work completely unchanged while sharply increasing peak speed and peak power — exactly what the faster-lift experiment demonstrates. Electrical input is mechanical work divided by drive efficiency, with the difference between them counted as drive loss: at 80% efficiency the baseline lift's 294.3 J of mechanical work requires 367.875 J of electrical input, leaving 73.575 J of loss, while setting efficiency to 100% makes electrical input exactly equal mechanical work with zero modeled loss. The ideal holding brake draws zero power once the load is stationary, since power is force times velocity and velocity is zero at rest — even though the cable must still supply full tension to support the load's weight.

Frequently asked questions

Is cable work during a lift always just the load’s weight times height (mgh)?

Only once the lift is finished and the load is back at rest. Mid-lift, the moving load also carries kinetic energy, so cable work at any instant equals mgh + ½mv² — using mgh alone during acceleration or deceleration would understate the actual work being done.

Why does the ideal holding brake use zero power to keep the load suspended?

Because mechanical power is force multiplied by velocity, and the load’s velocity is exactly zero once it comes to rest at the top of the lift. The cable still must supply full tension to support the load’s weight — that has not changed — but with zero velocity, zero power is needed to maintain that static condition in this ideal model.

If you lift the same load the same height faster, does that change the total work done?

No. Final mechanical work depends only on the completed height and load mass (mgh once the load returns to rest), not on how quickly the lift happens. What changes with a faster lift is the peak speed and peak power required during the motion — the faster-lift experiment confirms the same 294.3 J of final work whether the lift takes 6 s or 3 s.

What does this hoist energy model leave out?

It is a prescribed smooth rest-to-rest motion profile, not a motor torque-control simulation, using a massless inextensible rope and a frictionless fixed sheave with no gear or drum inertia. Drive efficiency is held constant, there is no lowering or regenerative braking, and the height/duration ranges are limited to keep cable tension positive throughout the entire lift.

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