Power Output of a Motor PhysicsP = F·v for Constant-Speed Vertical Lift
Introduction
Mechanical power is the rate at which a force does work. When a motor lifts a block at constant speed, the tension in the rope equals the block's weight throughout the ascent. Power is therefore fixed at P = F·v = m·g·v for the entire lift: a single number that depends on the load and the chosen speed, not on the height or the elapsed time.
The distinction between power and energy matters in every motor-sizing problem an engineer encounters. A motor moving a 10 kg block to a height of 8 m transfers exactly 784.8 J of gravitational potential energy regardless of whether the trip takes 2 seconds or 20 seconds. What changes with speed is the rate of energy delivery, and therefore the size of motor required.
Most people expect that lifting heavier loads to greater heights demands more power. That expectation is partially correct but incomplete: height controls only how long the motor runs and how much total work is done, not the instantaneous power. With mass at 10 kg and speed at 3 m/s, the Power readout holds at 294.3 W whether the target height is set to 2 m or 20 m, a result the W(t) and P(t) traces in the simulator make immediately visible.
The Physics Explained
The sim constrains the motor to lift at constant speed throughout the ascent. That constraint removes acceleration from the problem: the net force on the block is zero, so the upward tension from the rope equals m·g exactly. No surplus force exists to speed the block up, and no deficit exists to slow it down. Power is therefore P = tension × velocity = m·g·v, a product of two slider values that never changes during a run. With mass at 10 kg and speed at 3 m/s, the Force readout shows 98.1 N and the Power readout shows 294.3 W from the first frame to the last.
Work accumulates differently. At any moment during the lift, the work done on the block equals the force multiplied by the distance already covered: W = m·g·y, where y is the block's current height. Since y grows at rate v, the W(t) trace on the right panel is a straight line with slope P. The amber W(t) line in the graph rises at exactly 294.3 J per second when mass is 10 kg and speed is 3 m/s, and it terminates at W = m·g·h = 10 × 9.81 × 8 = 784.8 J when the block reaches the 8 m target. The P(t) line, drawn in forest green, is flat throughout because power is constant: the motor neither speeds up nor slows down its energy delivery rate.
Doubling either slider input doubles the power output through a simple proportionality. Raising mass from 10 kg to 20 kg at speed 3 m/s takes the Power readout from 294.3 W to 588.6 W, confirming the linear scaling with load. Raising speed from 3 m/s to 6 m/s at mass 10 kg produces the same jump: 294.3 W to 588.6 W. The time to complete the lift changes only when speed changes: at 3 m/s the 8 m trip takes 8/3 ≈ 2.67 s; at 6 m/s it takes 1.33 s. Doubled mass at unchanged speed leaves the time at 2.67 s while doubling the work done to 1569.6 J, so the higher-power motor is running for the same duration but delivering energy twice as fast.
At the slider extremes the numbers scale predictably. With mass at 50 kg and speed at 10 m/s, the Power readout reaches 4905 W and the total work over the 20 m maximum height is 9810 J, completed in exactly 2.0 s. The cross-check P × t = 4905 × 2.0 = 9810 J = W confirms that the energy and power figures are internally consistent throughout the slider range.
Key Equations
With mass = 10 kg, speed = 3 m/s, and g = 9.81 m/s²: P = 10 × 9.81 × 3 = 294.3 W. The Power readout in the simulator displays 294.3 throughout the lift at these slider settings. Because the speed is constant, P does not vary with time, which is why the P(t) graph on the right panel is a flat horizontal line rather than a curve.
At constant speed the net force is zero, so the motor's upward pull exactly equals the gravitational pull downward. With mass = 10 kg: F = 10 × 9.81 = 98.1 N. The Force readout confirms this value at the default slider settings. Increasing mass to 20 kg raises F to 196.2 N and simultaneously doubles the Power readout to 588.6 W at unchanged speed, because P = F · v and F has doubled.
Work depends on height and mass, not on speed. With mass = 10 kg and height = 8 m: W = 10 × 9.81 × 8 = 784.8 J. The Work Done readout reaches 784.8 J when the block arrives at the target, regardless of whether speed is set to 0.5 m/s or 10 m/s. Changing only the height slider from 8 m to 20 m raises the final work to 1962 J while leaving the Power readout unchanged at 294.3 W.
At mass = 10 kg, speed = 3 m/s, and height = 8 m: t = 8 / 3 ≈ 2.667 s. The Time readout in the simulator terminates near 2.67 s at these settings. As a consistency check, P × t = 294.3 × 2.667 ≈ 784.8 J, matching the Work Done readout exactly. Reducing speed to 1 m/s at unchanged mass and height extends the run to 8.0 s and lowers the Power readout to 98.1 W, but the final Work Done remains 784.8 J because the same mass has moved through the same height.
Key Variables
| Symbol | Name | Unit | Meaning |
|---|---|---|---|
| P | Mechanical power | W | Rate of energy delivery by the motor; constant during a fixed-speed lift |
| F | Lift force | N | Upward tension equal to m·g at constant speed |
| v | Lift speed | m/s | Constant upward velocity of the block set by the speed slider |
| m | Mass | kg | Mass of the block being lifted |
| g | Gravitational acceleration | m/s² | 9.81 m/s², fixed in this simulation |
| h | Target height | m | Vertical distance the block travels; sets total work done but not power |
| W | Work done | J | Total energy transferred to the load: m·g·h |
| t | Lift time | s | Duration of the ascent: h/v |
Real World Examples
How do elevator engineers size a motor for a given load?
An elevator motor must continuously overcome gravity on the car, passengers, and counterweight net load while moving the cabin at its rated speed. The required power follows directly from P = m·g·v: a net moving mass of 1000 kg lifted at 1.5 m/s demands at least 1000 × 9.81 × 1.5 = 14 715 W of mechanical output, to which engineers add efficiency margins of 15 to 25 percent for gearbox and rope losses before specifying the motor's electrical rating.
Lift speed is the dominant design lever: doubling the cabin speed from 1.5 m/s to 3 m/s doubles the required mechanical power at unchanged load, which is exactly the proportionality the simulator reproduces. With mass set to 10 kg and speed raised from 3 m/s to 6 m/s, the Power readout climbs from 294.3 W to 588.6 W, a clean 2× ratio that confirms the linear dependence on v.
The time to lift the same 8 m target halves from about 2.67 s to about 1.33 s, while the Work readout holds at 784.8 J because the energy to move the load through a fixed height is independent of how fast the trip is made. Elevator specifications therefore list two separate figures: motor power (which determines how fast the cabin can travel at rated load) and total energy per trip (which determines operating cost over a service day).
Why does a crane's rated power limit both load and hoisting speed?
A crane's motor delivers a fixed ceiling of mechanical power. Since P = m·g·v, that ceiling sets a rigid trade-off: lift a heavier load and the maximum safe hoisting speed drops in exact proportion, or hoist faster and the permissible load decreases. A crane rated at 50 000 W can lift a 5000 kg load at a maximum of 50 000 / (5000 × 9.81) ≈ 1.02 m/s, or a 500 kg load at up to 10.19 m/s.
The simulator makes this trade-off concrete. With mass at 50 kg and speed at 10 m/s, the Power readout reaches 4905 W, the slider-maximum scenario. Halving speed to 5 m/s at the same mass drops power to 2452.5 W while the Work Done readout still converges to the same 9810 J once the block reaches 20 m, because the height and weight, not the speed, set the total energy transferred.
This is why crane load charts list both a rated lift capacity and a maximum line speed: exceeding either individually can stay within the power envelope, but demanding both simultaneously overloads the motor. Site engineers reading load charts are applying P = m·g·v implicitly every time they pick a hook configuration for a given pour height and schedule.
How does a cyclist's power output determine climbing speed on a steep hill?
On a steep gradient where aerodynamic drag is small relative to the gravity component, a cyclist's climbing speed is governed by the same equation as the motor sim: v = P / (m·g). A rider producing 300 W and carrying a combined rider-plus-bike mass of 75 kg climbs at 300 / (75 × 9.81) ≈ 0.408 m/s on a fully vertical wall, or about 4.1 km/h on a 10 percent grade where the vertical component of velocity is one-tenth the horizontal speed.
Professional climbers targeting alpine cols at race pace sustain 5 to 6 W/kg, which translates to roughly 350 to 420 W for a 70 kg athlete and predicts ascent rates that match observed race data closely. The simulator captures the core relationship: setting mass to 10 kg and speed to 3 m/s puts the Power readout at 294.3 W. Keeping power fixed while cutting mass to 5 kg would require speed to rise to 6 m/s to conserve P = m·g·v, exactly how a lighter rider travels faster up the same hill at the same sustainable power output.
Sports scientists use this proportionality to evaluate training interventions: a 2 percent reduction in body mass at constant power output produces a 2 percent gain in climbing speed, a marginal gain that compounds across a multi-hour mountain stage and explains the extreme weight management strategies visible at the professional level.
Further Reading
- Net work with friction: extending the work-energy theorem to cases where friction opposes motion.
- Gravitational potential energy on a hill: the stored-energy view of the same vertical-lift scenario.
- Work done by a variable force: how the calculation generalises beyond the constant-speed lift case.