Problem — Power of a lift

Problem

A lift of total mass $500\;\mathrm{kg}$ (cabin + passengers) rises at a constant speed of $1{,}2\;\mathrm{m/s}$. Calculate the mechanical power delivered by the motor, neglecting friction.

Solution

At constant speed the acceleration is zero, so the motor’s force exactly balances the weight: $F = m\,g$. Mechanical power is force times speed: $P = F\,v = m\,g\,v$ Substituting the data: $P = 500\;\mathrm{kg}\cdot 9{,}8\;\mathrm{m/s^2}\cdot 1{,}2\;\mathrm{m/s} = 5880\;\mathrm{W}$ $\ev{P \approx 5{,}9\cdot 10^3\;\mathrm{W}}$

That is roughly $5{,}9\;\mathrm{kW}$: each second the motor must do the work needed to lift $500\;\mathrm{kg}$ by $1{,}2\;\mathrm{m}$.

Links

Topics: Work and energy Concepts: Power · Work done by a force Skills: Symbolic setup Objects: Lift