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## Homework Statement

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?

## Homework Equations

## The Attempt at a Solution

1.

w=m*a*s+637,65 N*s

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

2.

w=m*a*s+637.65*s

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?

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