5

I am trying to figure out the amount of work done when I walk X miles or for X minutes. So I got Work=Force x Distance and Force=Mass x Acceleration and Acceleration=(change in velocity)/time. I am stuck at the acceleration part. As far as I am concerned my velocity is constant, so Acceleration=0 therefore Force and Work are 0. This doesn't seem right. What am I missing here?

Qmechanic
  • 201,751
Dan
  • 183
  • 1
  • 6
  • 3
    It's because you have oversimplified the problem. In your explanation are effectively talking about a blob sliding on a flat, frictionless surface at constant velocity. The problem is far more complicated. – Will Jun 06 '13 at 16:24
  • 1
    Ok I got you. This is still oversimplified, but does it even make sense. Let Acceleration = (avg. velocity) / (1 sec). The thought behind this is every time you take a step your almost starting from a stop since your momentum is minimal. So each step you accelerate to your average velocity and a step takes about a second. – Dan Jun 06 '13 at 16:36
  • 2
    Think for example about lifting your leg, does this require work? Dan08, what do you think? – Will Jun 06 '13 at 16:39
  • 2
    You confusion stems from two thing (1) the distinction between "physics work" and the colloquial "man, this is hard" meaning and (2) while the net force is zero your muscles must constantly provide a force to counter losses to friction in various guises. – dmckee --- ex-moderator kitten Jun 06 '13 at 20:36
  • 2
    This "The thought behind this is every time you take a step your almost starting from a stop since your momentum is minimal." is not a good reflection of human locomotion which makes good use of momentum carried from step to step. – dmckee --- ex-moderator kitten Jun 06 '13 at 20:38
  • It's usually best to think of work as being defined as the mechanical transfer of energy. By this definition, it should be clear that your foot's force on the pavement isn't doing work (the pavement doesn't gain or lose energy), but internal forces in your body are doing work (e.g., your leg gains and loses energy throughout your stride cycle). –  Jun 06 '13 at 21:53
  • 1
    @BenCrowell, sorry to be picky but the wear and tear on old steps show at least a little work is being done. – Nic Jun 07 '13 at 16:16

2 Answers2

6

When it comes to modeling the human body and its movements it becomes a lot more complex, because you actually have to model the different components of your body and how they attribute to the system. Plus you are not traveling at a perfectly constant velocity when walking at a steady pace.

If you were to model the work done by the human body while walking, you obviously will consider more of the lower body. I would personally neglect what the upper body does, even though your back muscles help move the legs and arms help stability and the respiratory system. However you want to know the quantity of work done by your legs, essentially.

Essentially (and crudely) it would be modeled by a set of joints that represent your legs and a block of mass M (your weight in total), and there it just a series of statics problems where you may model the movement of S number of steps and you can quantify the work done by your legs per step, and don't forget friction.

signus
  • 582
-1

When we walk 3 forces act on us normal, gravitational and friction. For normal and gravitation force perpendicular to total displacement so by dot product rule work done is 0.and for static friction 0 displacement, when we displace we don't contact with ground so friction don't do any work.so over all work done in walking is 0.

Hpb
  • 1