Q.A lift motor raises a total load (lift cabin plus passengers) of mass $1500\ \text{kg}$ through a height of $20\ \text{m}$ in $25\ \text{s}$, moving at constant speed throughout, while also overcoming a constant frictional resistance of $500\ \text{N}$ in its guide rails. Find

1 Horsepower (HP) = 746 watt.

Horsepower is a unit of power historically used to compare the output of steam engines to that of draft horses; it remains in common use for engines and motors. Its exact SI equivalent (mechanical horsepower) is defined as 1 HP = 746 W (sometimes stated more pre …

Power is the rate of doing work; for a force F producing velocity v, P = Fv.

Work done by a constant force F over a small displacement ds is dW = F·ds. Power is the rate of doing this work:

P = dW/dt = F·(ds/dt) = F·v

This also checks out dimensionally: [Power] = [Work]/[Time] = (M L^2 T^-2)/T = M L^2 T^-3. And [F][v] = (M L T^-2)(L T^-1) = M L^2 T^-3 — matches. An option like Fv^2 would have dimensions M L^3 T^-4, which is NOT power, so it cannot be right regardless of the printing issue.

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The scalar (dot) product of force and velocity equals instantaneous power, $P = \vec{F}\cdot\vec{v}$.

Work done by a force over a small displacement is $dW = \vec{F}\cdot d\vec{s}$. Dividing both sides by the time interval $dt$:

$\dfrac{dW}{dt} = \vec{F}\cdot\dfrac{d\vec{s}}{dt} = \vec{F}\cdot\vec{v}$

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Power is the scalar (dot) product of the force vector and the instantaneous velocity vector: P = F→ · V→ = FV cos θ, where θ is the angle between them.

Power is defined as the rate at which work is done: P = dW/dt. For a small displacement dr→, the work done is dW = F→ · dr→ (only the component of force along the displacement does work). Dividing by dt:

P = dW/dt = F→ · (dr→/dt) = F→ · V→

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