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Homework Help: Work done on moving coin.

  1. May 31, 2010 #1
    1. The problem statement, all variables and given/known data

    Supposing you have these two scenarios:
    String is attached to a coin.
    (A) You raise the coin vertically from a horizontal table through a distance of 10cm.
    (B) Pull the coin up a smooth book that is tilted with respect to the table until its vertical distance changes by 10cm.

    The coin moves at constant speed in both these cases.

    2. Relevant equations
    This is supposed to be theory based.

    3. The attempt at a solution
    I would like to know whether there is more work done in case B, because the coin moves a greater distance, since it is basically moving up the hypotenuse?....I understand that W=Fd, and hence the greater distance travelled. However since F=ma, and it is moving at constant speed, hence a=0, so there is no force?

    When i tried this out, it felt the same to me in both cases.

    I was told, was that they would be the same because of "the conservation of energy"...but i don't really understand that
  2. jcsd
  3. May 31, 2010 #2
    Oh dear... I dunno where to start.
    1. F=ma is always true, but not necessarily in this case. Consider a coin on this book without you touching it. It would slide down, wouldn't it? Now, the force you are applying, is needed to prevent the gravitation from pulling the coin down, so it doesn't equal 0. It's actually eqal m*g*sin(alpha), alpha being an angle between the book and the table.
    2. Now, the work is not simply Fd, it's F*d*cos(angle between force and distance (again, lack of vocabulary is disturbing :/ )), but you can assume that you are pushing it parallelly, so that you get cos(0)=1. So your work W=F*d. But again, we have calculated the force in point 1. So we get W=m*g*d*sin(alpha). Now, think what is d*sin(alpha) and everything should be clear :P
  4. May 31, 2010 #3
    Hey, this is actually for one of my undergrad courses titled "The magic of Physics", which is for students who have never really done physics. So we don't do any math in it. Lol, so is there anyway you could explain your second point in basic terms pleaasse?...lol
  5. May 31, 2010 #4
    have you been taught about gravitational potential energy? that theory would allow you to explain this problem easily without messing with any math whereas using the base equation w=fd seems to require the math to understand this problem.

    If you have, notice that work done by gravitational forces are conservative. That means you can move from point a to point b by using any path and the work done by the force of gravity will be the same. If you are raising an object, you do work to overcome gravity's work. So gravity would have it that all objects rest low, meaning it works to bring objects down. You must do exactly the same work that gravity wants to do when bringing an object down to bring it up. Therefore, your contributions of work when raising an object are also conservative.
  6. May 31, 2010 #5
    Well, let's try. Imagine a frictionless...something..let's say ice. And imagine you are moving the coin up the slope made of ice ;). No, back. Imagine you are moving it on a completely horizontal ice. It doesn't require any force of you to do it, since there is no friction ,so that the coin is moving completeley freely, right?
    Now, back to the slope. Now, our moving up the slope can be divided into two parts-you are moving the coin horizontally and veritcally. As we've already stated, the horizontal moving doesn't require any force. And so, we need the force only to move the coin up vertically. And this force, of course, is not the same as if you were moving the coin up without any slope/book/whatever, BUT! the distance you have to use it on is longer then the one simply needed to lift the coin. And so, mathematically it comes ou, the the result of the multiplication force and distance in each situation (lifting it up and pushing it up the slope) is exactly the same.

    You could also consider the work as a change of an energy. In this contest we can concentrate on two types of energy only-kinetic and potential. Let's say, that on the table the coin has the potential energy=0 (and this is the assumption we can always make) and kinetic energy=0 as well (since it's not moving). Now, lift the coin up 10 cm. After you've moved it, it's no longer in the move, so the kinetic energy is equal to 0 again. Now, the potential energy, however, has increased, since the height of the coin above the table is no longer 0 (it's no longer on the table). But in both cases the distance is the same, and so is the change of energy, and this is what the work actually is, the change of energy.
  7. May 31, 2010 #6
    Even on a frictionless surface, force is needed to begin motion, but no force is needed to continue motion. Yes, what you said is true for this problem due to velocity's constancy, but what you said was not truest.
  8. May 31, 2010 #7
    You are absolutely right, but I didn't want to make things more complicated. There is a silent assumption made in this exercise, that you can start moving something with a constant velocity (no acceleration), which is obviously impossible, but since those velocities can be recahed in almoste no-time, i guess we are quite right to do so.

    Anyway, my deepest apologies for not being exact.
  9. May 31, 2010 #8
    Irycio- So in both cases the KE while the coin is moving diagonally or vertically is the same right?....hence the PE is the same...

    xcvxcvvc- I read up on gravitational potential and it makes a lot of sense, thank you for bringing that to my attention.
  10. May 31, 2010 #9
    Doesn't matter, you are only interested in the initial and final state, where it's equal to 0. You could actually speed the coin up to 1000 m/s, applying a work to it, but it would have to be stopped after that, and would "give the work back", so that they would cancel each other.
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