(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. 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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Work - Acceleration of an elevator

**Physics Forums | Science Articles, Homework Help, Discussion**