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I Work and displacement

  1. Apr 26, 2016 #1
    Work = force x displacement, so if there is zero displacement, the work is zero, right? But if I get up and walk across the room and walk back again to my starting place, the net displacement is zero, however I am told that there is still work performed. The explanation given for this is that work is a scalar quantity even though displacement is a vector. I'm not sure why this means that the net displacement needs to be broken into its component parts (effectively treating it as the scalar quantity of distance) in order to calculate the work performed...?
     
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  3. Apr 27, 2016 #2

    cnh1995

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    Work is the dot product of force and displacement vectors i.e. W=∫F.dx. or |W|=F*x*cosθ. So, it is a scalar quantity.
     
  4. Apr 27, 2016 #3
    Again, I'm not sure why this means that displacement should be treated as a scalar when calculating work. Please clarify this for me if you can.
     
  5. Apr 27, 2016 #4

    cnh1995

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    It shouldn't. Force and displacement are both vectors. Their dot product i.e. 'work' is a scalar quantity. All the dot products give scalar quantities. Hence, dot product is also called 'scalar product'.
    This means work is the product of displacement and component of force in the direction of displacement.
     
  6. Apr 27, 2016 #5
    Can you explain this in the context of my example of walking across the room and back? Is the displacement zero, or isn't it? Is it realistic to talk about work in this example, or does the (apparent) zero quantity of displacement mean there is zero work performed?
     
  7. Apr 27, 2016 #6

    cnh1995

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    Suppose you pushed a box from A to B with some force F. Work done by you in this process is F*d. If you pushed the block back to A, work done is again F*d. Note that while pushing the block back from B to A, both force and displacement changed their directions, so the work is not 0. If this were a conservative field, work done would be 0.
     
  8. Apr 27, 2016 #7
    What is a conservative field?
     
  9. Apr 27, 2016 #8

    cnh1995

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    Conservative field is a field in which work done is path independent. No matter how you go from A to B, work done is same along all the paths. Examples are gravitational field and electrostatic field. If you threw a ball up in the air with some veocity, it will go up and stop at some point. Then it will start falling and when it reaches the ground, it will have the same velocity as it had when it was thrown up. This means, work done on the ball is 0. Here, displacement of the ball changes its direction but force doesn't. Gravitational force always pulls the ball downward. Going up, the ball loses energy and falling back, it gains energy. Thus, net work done is 0.
     
  10. Apr 27, 2016 #9

    cnh1995

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    In your 'walking across the room' example, "net displacement" is 0. But work is not Force*net displacement. It is actually ∫F⋅dx i.e. summation of force*instantaneous displacement between two points. Also, force and displacement both change their directions when you walk back to where you started. Zero displacement does not necessarily mean zero work.
     
    Last edited: Apr 27, 2016
  11. Apr 27, 2016 #10
    Ok well I'm still far from being 100% clear on any of this but I appreciate your attempts to illuminate me. I'm new to studying physics and obviously struggling to grasp basic concepts. Perhaps I'll eventually gather enough knowledge to understand the rationale behind those concepts. Thanks for your help cnh1995.
     
  12. Apr 27, 2016 #11
    So when you find the work done when u push the box from A to B and back, you really need to consider every moment.

    U start pushing the box. U push the box a nanometer. stop. find the work by taking the (dot) product of force and displacement. U again push the box 1 nanometer .stop. Again find the work done .you need to do this an infinite number of times to find the total work done by adding the individual work for each moment.

    Thus your work doesn't depend on the 'net displacement ' . Rather, it depends on the individual forces and displacements at every moment.
     
  13. Apr 27, 2016 #12

    cnh1995

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    You are welcome!
    When work is done, there is always change in energy of the body. When you move across the room and come back, your energy is lost. You do not gain any energy in this process. This means some non-zero work has been done. When work done is zero, initial and final energies of the body must be equal. If a body loses energy when going from A to B but gains the lost energy while coming back from B to A, work done is zero. Because here, initial energy and final energy are equal. That's what I was trying to say in #8 using the example of a ball thrown up against gravity. Going up, it loses energy and coming back, it regains energy. This happens because force doesn't change its direction, only displacement does. Hence, when the ball is going up, force of gravity is resisting its motion and the ball is losing energy. When the ball is falling down, the same force of gravity now aids its motion, hence, the ball gains energy.
    Good luck with your studies!:smile:
     
  14. Apr 27, 2016 #13
    for every step take take friction is always pushes you forward , so instantaneous displacement is in dir of instantaneous force.
    so instantaneous work is +ve , so net work by friction on body is non-zero
     
  15. Apr 27, 2016 #14
    Unfortunately you picked up a quite complex system, that of a man walking. This does not help your understanding.
    Who told you that the work is not zero and in what context?

    You can calculate work as force time overall displacement only if the force is constant. And constant means both in magnitude and direction. Otherwise you nedd to use the integral form.
    If you apply a constant force to an object moving from A to B, and then at B you switch the direction of the force at B and apply the force from B to A, the force is not the same. You need to integrate (or here, just to split into two). And the work done by this (or these two) force(s) is not zero.
    If the direction of the force is not switched at B, then the work done by that force on the ABA path is indeed zero.
    Context is important.

    All this does not apply in a simple way to walking. Even if you stand still, your muscles do some work and you spend energy. It's better to start with simpler systems than a living being.
     
  16. Apr 27, 2016 #15
    Ok that's the critical information that I was missing. So if the force is changing you need to calculate the sum of all the forces and multiply that by the sum of all the displacements produced by those forces?

    How does this relate to the explanation in terms of dot products and cross products?

    The context was a biomechanics prac class. Hence the use of a living system as an example.
     
  17. Apr 27, 2016 #16
    No, you don't add separately.
    You add the dot products of force and displacements. Which for continuously varying force means do an integral.
     
  18. Apr 27, 2016 #17
    I don't know what 'do an integral' means. Sounds like a skateboard trick of some sort.
     
  19. Apr 27, 2016 #18
    No, it is not a skateboard trick but a calculus one. :)

    But until you learn some more math you can divide the motion into portions so that the force is constant for each one and add the dot products (force times displacement) to get the total work.
    An integral is something like this but where the portions are very small (infinitesimal).
     
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