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

Work done

  1. Jan 21, 2004 #1
    I have a simple question.

    If say for distance x, a person needs to go and come back at the intial position,he has 2 choices 1.To get up, go and come back on his seat. 2.To drag the chair he is sitting on and pull back on his intial position with his seat itself.
    In which case is the work done or energy utilised by him more.What's the better thing to do.
  2. jcsd
  3. Jan 21, 2004 #2
    It's the same in both cases, I beleive. In the first, the work done equals F*(s_2 - s_1). In the second, the man can never be sure if he's moving towards the chair or the chair towards him if it/he is moving at a constant velocity, so it can be treated exacly like the first example. Work then equals F*(s_2 - s_1). In both cases s_2 = s_1 since he ends up where he started, in which case the work done is zero.
  4. Jan 21, 2004 #3


    User Avatar
    Homework Helper

    Probably case 2. 1 will most likely incur some friction. If 1 does not, then how would the chair be moved? In 2, the only lossy friction could be sliding up out of and down into the seat.
  5. Jan 21, 2004 #4
    If it doesn't take more energy to move an object with yourself as opposed to just moving yourself, well, physics get's redefined.
  6. Jan 22, 2004 #5
    Thanks a ton Pallidin..!!
    But actually its not so easy to redefine Physics.

    According to my understanding the work done in case 1 is :a)getting up and sitting back on chair b) walking up and down the x distance.
    In case 2 is :- The force applied to drag the chair as well as moving it along the x distance up and down.

    In case 1 - we are working against the gravitation.
    In case 2 - we are working against the friction.

    Though in moving up and down as the distance covered in direction of force,i feel the work done would be zero as the displacement is zero.
    Though i too agree with you that case 2 might be contributing to a higher work done but can we actually overlook the work done against Gravity while the man gets up and sits again on his seat.

    Do let me know if u feel i am wrong.
  7. Jan 22, 2004 #6


    User Avatar
    Homework Helper

    You need to decide what exactly you mean by work: input or output.

    The last sentence I quoted here suggests output. But everything before that suggests input. Further more, in case 1, you are only working against gravity when you get up; you are working with it when you sit down, so this would cancel in a simplified model. You always have to work against friction.

    One other point: you do not necessarilly have to do work to move a distance x. A human being consumes energy to sustain life, and energy is consumed by the muscles when force is exerted because the fibers are contracting and relaxing periodically during the exertion (to a lessened degree in a toned muscle, but still present). So, do you want to count this input energy as work? And do you want to count the energy required to stay alive as well as the energy require to exert force (i.e. move a body part)? In order to answer the question, you need to first answer what you mean by work.
  8. Jan 22, 2004 #7
    Mithus, I think I understand your question alot better than I did when I posted. Interesting, I guess I never thought about it in the way you described, but you do have a point.
    Turin brought up good points too. I guess I'll just listen to see how this works out.
  9. Jan 22, 2004 #8


    User Avatar
    Homework Helper

    I read through the posts again. I realized an error in my first post. I referred to case 1 when I meant case 2, and vice versa.
  10. Jan 29, 2004 #9
    When you stand up, you do work against gravity. The work done is equal to your change in gravitational potential energy, V=mgh. If you have a mass of 100 kg and the position of your center of mass changes from 50cm to 100cm when you stand, and using [tex]g = 10m/(s^2)[/tex], then standing up requires 500 Joules. Incidentally, that's about 100 little-c calories or 0.1 big-C Calories (the kind in food).

    When you scoot yourself in the chair, you're doing work against friction. Assuming we have a flat surface, the formula for the work done is W = F*d = (umg)*d

    u (should be a mu) is the coefficient of frict.

    m is your mass, plus the negligible mass of the chair (so still 100kg)

    g is still 10 m/s^2

    d is how far you're going.

    The coefficient of friction between rubber (I assume little rubber footies on the chair legs) and a solid can be somewhere between 1 and 4, roughly--this is a dimensionless quantity--so assuming you're dragging the chair three meters, it could take somewhere between 3000 and 12000 Joules of work (about 1 to 3 Calories burned)

    But the chair could easily have a coefficient of (kinetic) friction of much less than one: teflon on teflon is 0.04; a lubricated metal-on-metal contact can get 0.06. In those cases, it would only take 120 to 180 Joules, which is less than the energy required to stand up.

    As other posters have remarked, these calculations do not take efficiency into account: it *requires* 500 Joules of work to stand up; that doesn't mean that you didn't spend more than that. And you certainly couldn't drag with perfect efficiency on a slippery floor.

    So it mostly depends on the coefficient of friction and how far you're going. If the chair has little rubber nubs, scooting only make sense for very short distances (order of a few centimeters). If the chair has wheels, then it makes sense to scoot around in it for all but the longest trips (over a few tens of meters.) Since that agrees with my personal intuition and past experience with chair-scooting, I'll stand by it.

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