# Work - Acceleration of an elevator

1. Dec 3, 2006

### tobias.mueller2

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?

Last edited: Dec 3, 2006
2. Dec 3, 2006

### Astronuc

Staff Emeritus
The equations are correct, but the acceleration derivation is incorrect.

The work equivalent to the change in kinetic energy. The elevator starts at rest (KE = 0) and accelerates to a new KE (0.5* m * v2).

The distance of 1.5m is correct, and the contribution of friction (637,65 N*1.5 m) is correct.

In the first part, does one need to consider the change in graviational potential energy, mgh or mgs, where s is the vertical distance traveled (= h)? This is considered in the second part.

Work/energy = force * distance, assuming constant force, otherwise if F = F(x), then one must integrate $\int{F(x)}\,dx$

3. Dec 3, 2006

### tobias.mueller2

w=∆e[kyn]+∆e[pot]+637,65 N*1.5 m=0.5*m*v2-0+m*g*s-0+637,65 N*1.5 m=0.5*1300 kg*1.5 m/s2+1300 kg*9.81 m/s2*1.5 m+637,65 N*1.5 m=21548,475 Joule

In the end, I get the same result. Do I made any fault?