Talk:Work (physics)/Archives/2015/January

Conflict? Work/energy equivalence vs work against of done by gravity?
The article explains that if you apply a force over a distance, that's work. Apply the same force over twice the distance, and you've done twice the work; that seems pretty clear. So if you lift a weight a certain height above the ground, you've done a certain amount of work; if you lift it twice as high, you've done twice the work.

And the article says that if you drop an object a certain distance, gravity does a certain amount of work on the object; if you drop it twice the distance, gravity does twice as much work.

But then the article talks about the equivalence of work and energy.

If you drop the object twice a certain distance, it will be going twice as fast by the time it hits the ground (disregarding air resistance and the drop-off of gravity when the distance is large). That means it'll have four times the kinetic energy, because kinetic energy is 1/2mv2.

Can someone have the article sort this out? Or is it already done, and I'm just missing it? Uporządnicki (talk) 02:08, 23 January 2015 (UTC)


 * I think I've begun to realize the answer to my question, and I'm actually feeling a bit stupid now. I didn't want just to delete it because that would leave an odd trail of disembodied signatures.  So I decided to be honest and just strike it out.  In short, what I'm realizing is, dropping the body from twice the height won't double the speed.  To double the speed (and thus, quadruple the kinetic energy), you have to drop it from such a height that it falls for twice the length of time.  That's considerably more than twice the height.  In fact, it turns out to be four times the height. Uporządnicki (talk) 02:57, 23 January 2015 (UTC)
 * I agree that dropping the body from twice the height won't double the speed. But it will double the kinetic energy which is related to the work done; and work is force times distance (or force times height).
 * You say that to quadruple the kinetic energy you have to drop the body from such a height that it falls for twice the time. I disagree because time plays no part in determining the work done or the change in kinetic energy. To quadruple the kinetic energy requires four times as much work done by the weight of the body, and work is force times distance, so it requires the body to be dropped from four times the height.
 * One part of the work-energy theorem states that the change in kinetic energy of a body is equal to the work done by the resultant force acting on that body. When a body is falling under the influence of no force other than its own weight, the resultant force on the body is its weight. Dolphin  ( t ) 07:27, 23 January 2015 (UTC)
 * Four times the height IS twice the time. Yes, it's Force times the Distance.  But you can express the distance as a function of the time: Distance fallen = acceleration x time2.  To quadruple the kinetic energy (of ANY particular moving body), you double the speed.  To double the speed of a free-falling (or ANY constantly accelerating) body, you double the time falling.  When you double the time falling, you quadruple the distance.


 * In short, when you drop an object and it falls freely: Double TIME falling = double DISTANCE fallen FINAL SPEED = quadruple (i.e. double2) FINAL SPEED DISTANCE fallen = quadruple (i.e. double2) KINETIC ENERGY.


 * Notice, by the way: TRIPLE time falling = TRIPLE distance fallen final speed = NINE TIMES (i.e. triple2) final speed distance fallen = NINE TIMES (i.e. triple2) kinetic energy.Uporządnicki (talk) 15:28, 23 January 2015 (UTC)


 * The concept of work is very useful because of its wide applicability in all general situations, including applicability to bodies that are subject to varying forces and varying accelerations. Our article on work is written with the assumption that a body’s acceleration won’t be constant. Your statements are true but only in the special situation where a body is experiencing a constant acceleration.


 * If you find it easier to think about the time for which a force is acting you might find it more useful to think about the impulse given by the resultant force, causing a body to change its momentum. In this way your thoughts won’t be constrained to a constant acceleration.


 * In the situation where the resultant force is constant you could make the following statement: To double the speed (and thus double the momentum), you have to drop it so that it falls for twice the length of time. Dolphin  ( t ) 05:22, 24 January 2015 (UTC)


 * Actually, I see I still had it garbled (although I was getting close; I think my error came in the translation of thought to keyboard). I've made the corrections in my last statement--presented in such a way as to show where I was wrong and what is corrected.


 * As for your comment, well, yes! I was particularly addressing the case of lifting/dropping a given body to/from a particular height or twice that height.  So, yes, constant acceleration.  It wasn't that I particularly preferred thinking in terms of the time; it's that I'd confused it with the point of how far the body is falling.  I was thinking:  1)  Drop the body twice as far.  2)  It falls twice as long (DING, DING, DING, DING!  ERROR! ERROR!).  3) It finishes twice as fast (ERROR IN CONSEQUENCE OF STEP 2), 4) thus, finishing with four times the kinetic energy (DITTO).  And how could that be, if you've done twice the work lifting it that high in the first place?  In fact, 2) it will fall √2 times as long, 3) finish √2 times as fast, 4) thus finishing with 2 times the kinetic energy.  Problem solved. Uporządnicki (talk) 05:55, 24 January 2015 (UTC)