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Question: An elevator can carry a maximum load of\(1800\,kg\) \(\left( {elevator + passengers} \right)\)is mov...

An elevator can carry a maximum load of1800kg1800\,kg (elevator+passengers)\left( {elevator + passengers} \right)is moving up with a constant speed of2m/s2\,m/s. The frictional force opposing the motion is400N400\,N. Determine the minimum power delivered by

Explanation

Solution

The mathematical equation to calculate power is P=F.VP = \mathop F\limits^ \to .\mathop V\limits^ \to where FFforce is and VVis velocity.

Complete step by step answer:
An elevator can carry a maximum load.
m=1800kgm = 1800\,kg
m\therefore \,\,mis mass including elevator and passenger.
Hence the elevator is moving upward with constant speed of2m/s2m/s.
V=2m/sV = 2m/s
V\therefore \,\,Vis constant speed in upward direction.
Frictional force opposing the motion is
f=400Nf = 400\,N
f\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+mgF = f + mg … (i)
g\therefore \,\,gis acceleration due to gravity g=9.8m/s2g = 9.8\,\,m/{s^2}
To compute total retarding force use given value in equation (i).
F=400+(1800)(9.8)F = 400 + \left( {1800} \right)\left( {9.8} \right)
=400+17640= 400 + 17640
F=18040NF = 18040\,N … (ii)
Equation to compute the power required to move the load with constant speed in upward direction is
P=F.VP = F.V … (iii)
Put the values of VVandFFin equation (ii).
P=(18040)(2)P = \left( {18040} \right)\,\left( 2 \right)
P=36080wattP = 36080\,watt
Hence total power required to move the load of 1800kg1800\,kgin upward direction with constant speed is36080watt36080\,\,watt.

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