Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work - Acceleration of an elevator

  1. Dec 3, 2006 #1
    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
    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

    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. jcsd
  3. Dec 3, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    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 [itex]\int{F(x)}\,dx[/itex]
  4. Dec 3, 2006 #3

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook