An elevator can carry a maximum load of$1800\,kg$$\left( {elevator + passengers} \right)$is moving up with a constant speed of$2\,m/s$. The frictional force opposing the motion is$400\,N$. Determine the minimum power delivered by

Hint: The mathematical equation to calculate power is $P = \mathop F\limits^ \to .\mathop V\limits^ \to $ where $F$force is and $V$is velocity.

Complete step by step answer:
An elevator can carry a maximum load.
$m = 1800\,kg$
$\therefore \,\,m$is mass including elevator and passenger.
Hence the elevator is moving upward with constant speed of$2m/s$.
$V = 2m/s$
$\therefore \,\,V$is constant speed in upward direction.
Frictional force opposing the motion is
$f = 400\,N$
$\therefore \,f\,$is frictional force
Hence the elevator is moving in an upward direction so motion is against gravity so gravitational force is also retarding the motion. So the total retarding force which opposes the motion is including the weight of the elevator and passenger.
Hence total retarding force is
$F = f + mg$ … (i)
$\therefore \,\,g$is acceleration due to gravity $g = 9.8\,\,m/{s^2}$
To compute total retarding force use given value in equation (i).
$F = 400 + \left( {1800} \right)\left( {9.8} \right)$
$ = 400 + 17640$
$F = 18040\,N$ … (ii)
Equation to compute the power required to move the load with constant speed in upward direction is
$P = F.V$ … (iii)
Put the values of $V$and$F$in equation (ii).
$P = \left( {18040} \right)\,\left( 2 \right)$
$P = 36080\,watt$
Hence total power required to move the load of $1800\,kg$in upward direction with constant speed is$36080\,\,watt$.

Note:
Power is scalar quantity which is computed by dot product of two vector quantity force and velocity. And its unit is watt.