An elevator with 1300 kg accrlerate from 0 m*s^-1 to 1.5 m*s^-1 in 2 s. After that, he drives further with a constant velocity of 1.5 m*s^-1. One must add a frictional force of 637,65 N.
The questions are:
1. How much work is needed to accelerate the elevator?
2. How much work is needed for the driving of the elevator with a constant velocity in dependence of the height?
The Attempt at a Solution
a=v*t^-1+9,81 m*s^-2=1.5 m*s^-1*(2 s)^-1+9,81 m*s^-2=10,56 m*s^-2
s=0,5*a*t^2=0,5*1.5 m*s^-1*(2 s)^-1*(2 s)^2=1.5 m
w=1300 kg*10,56 m*s^-2*1.5 m+637,65 N*1.5 m=21548.5 Joule
w=1300 kg*9,81 m*s^-2*s+637.65*s
I guess 21548 Joule is a way too many for just the acceleration of an elevator, isn't it?