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Work notion

  1. Jul 13, 2013 #1
    Hello, I have trouble to get what work is. I find physics notions difficult to understand and
    I'm kinda bit astray like Rainman in that movie when he walks that moving walkway the wrong way, realizing he's not progressing. Could you, please, explain me with human words what work really represent?

    W = ||F|| . cos(a) . d

    Knowing that "d" value will be affected by the inertness of the mass,
    should I consider "d" as a coefficient of efficiency?

    Is "d" supposed to be the distance covered by the body "after" the given force be applied on it because if so, in my intuition without any time factor taken into account, I hardly imagine any distance covered, time rules everything, moreover upon a perfect frictionless ice the body would never stop after the force be applied on so that "d" would be undefinite.

    Thanks for helping because physics formulas are for me as weird as multiplying apples with bananas and seeing watermelons come out.
     
    Last edited: Jul 13, 2013
  2. jcsd
  3. Jul 13, 2013 #2

    Dale

    Staff: Mentor

    Work is force times distance. Only the component of the force along the direction of displacement is considered (that is what the cos(a) factor represents).

    No. d is distance.

    No, d is the distance covered during the application of the force.
     
  4. Jul 13, 2013 #3

    Nugatory

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    Staff: Mentor

    d is the distance covered while the force is applied. When you stop applying force to the object on the frictionless surface it keeps on moving in straight line at a constant speed (no force, no friction, means no change in speed) with no additional work done on it.
     
  5. Jul 13, 2013 #4
    Hello, thanks for replies.
    "while the force is applied" ?
    so ||F|| applied for Δt causing to cover Δd...
    then why not? W = cos(a).||F||.Δt . Δd
    I still do not get it.
     
  6. Jul 13, 2013 #5

    WannabeNewton

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    Science Advisor

    Well for one that is not dimensionally consistent with the units for energy. Second, the ##\Delta d## itself implicitly encodes the information about the time of application because ##d = d(t)##. Work is defined, at a basic level, as ##F\cdot d##; explicit inclusion of a time variable would not have any physical meaning in the above sense.
     
  7. Jul 13, 2013 #6
    I give just one tug on a rope attached to a sled and this sled drifts over a distance d after the impulsion, then it stops due to friction, whatever, may I apply the formula F.d?
    or... is "d" the distance all along which I applied the force in the case I'd drag that sled, to say untill the sled cover distance "d" and over which it accelerates due to that very force? It's been ten years now I try to grasp the notion.
     
    Last edited: Jul 13, 2013
  8. Jul 13, 2013 #7

    jtbell

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    Using this distance d, you may apply W = F.d to calculate the work done by friction, assuming that (a) the frictional force F is constant, and (b) it acts over the entire distance d.

    If you want to calculate the work done by you, then you need to know (a) F = the amount of force that you exert during the tug, and (b) d = the distance that the sled moves during the tug.
     
  9. Jul 14, 2013 #8
    Thanks for your replies.

    If I apply "F" to something heavy, "d" will not be the same that if I apply "F" to something lighter, I can spend Joules and product null work therefore I consider work represent the amount of energy transmitted into motion? maybe..? I guess that's why work is expressed into Joules.

    How do I express the energy I release into Joules even if work product is null? because pushing the wall of my room I'd feel like I work too. force times time?

    My tug was supposed to be abrupt enough not to make any time factor come into account.
    Using the weight of the ropes, I don't know...

    Work quantifying cannot even be linear since the sled accelerates all over the distance I pull. And we quantify energy with that? At the end, and for a same given time, the amount of energy must increase really faster compared to the start though I do not spend more energy in pulling.

    I'm so lost with physics first concepts as you can see, I can apply them as well as others but I feel like i'm the only one wondering about those basic concepts, it doesn't surprise me that scientists end up to stumble across paradoxes. I got nothing. Has work been our standard to quantify energy. To say that the Joules of my cornflakes would product such amount of work in a system with perfect efficiency?
     
    Last edited: Jul 14, 2013
  10. Jul 14, 2013 #9

    Doc Al

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    No. If you apply a force but there is no displacement--such as when you push against a wall--you have done no mechanical work against the wall. You've transferred no mechanical energy into the wall. The work done by that force is zero.

    Since you are a biological system there is work being done--and chemical energy being consumed--within your muscles in order for you to maintain that force. Nonetheless, you are doing zero mechanical work against the wall.
     
  11. Jul 14, 2013 #10

    Nugatory

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    That actually makes the problem harder. I suggest that you may want to work through the case in which all forces are smoothly and steadily applied first; it's easier and tells you a lot more about the sudden tug case than the other way around. For the case of your sled...

    Calculus was invented to attack this and similar problems. But we can get by without calculus here. Suppose your sled has a mass of one kg and you apply a force of ten newtons to it for one second.

    Start with ##F=ma##, and we see that the sled accelerates at 10 m/sec2. We apply the force over one second, so if the sled started at rest it will be moving at 10 m/sec when we're done pushing. Its average speed during that one second is 5 m/sec (draw a graph of its speed as a function of time to see this - only works for a constant force and acceleration like we have here). If the sled moves at an average speed of 5 m/sec for one second, it will travel 5 meters during that second.

    Now we're good to go with ##W=Fd## - ten newtons over five meters gives us 50 Joules of work done on the sled.

    We can cross-check that result by calculating the kinetic energy of the sled when we're done pushing on it: ##E_k = \frac{mv^2}{2}## - v is 10, m is 1, and we get the same 50 Joules, the work done on the sled is now its kinetic energy.
     
    Last edited: Jul 14, 2013
  12. Jul 14, 2013 #11

    Dale

    Staff: Mentor

    As was already pointed out that formula wouldn't work because the units are wrong. However, the real reason is simply that it is defined the other way. There is no deeper reason for definitions. They simply are what they are, by definition.

    You are perfectly free to define a new quantity, e.g. the "lucas", which is defined the way you suggest. But that quantity will not be very useful for anything that I can think of.
     
  13. Jul 14, 2013 #12

    Dale

    Staff: Mentor

    You can always apply the formula as long as you know F and d. In this scenario you will do some positive work on the sled during the impulse and then friction will do some negative work on the sled. The two will be equal.

    I am not sure what you are asking. F is the actual force and d is the actual distance. I don't know the complication you are trying to explain.
     
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