Formula for Calculating WorkDate: 06/21/2005 at 18:07:59 From: Paul Subject: Calculus/physics definition of work and lifting My calculus text (Swokowski, Olnikc, Pence, 6th edition) gives the formula for work as W = Fd and then goes on to explain that if the force varies over the distance the formula becomes an integral. As part of an example, it then shows that the work to lift a 500 lb beam 30 feet would be 500 * 30 = 15,000 ft-lb. But don't we have to exert an upward force greater than the weight of the beam somewhere in our model in order to get the beam to move up? Then the work would be greater than 15,000 ft-lbs according to the definition of work. It seems that in setting up integrals or just using the formula W = Fd examples always use the weight of the increment or object to be lifted without taking into account the fact that more than that force must be exerted at some point in order for the thing to move upwards. By the formula W = Fd, if we greatly accelerate an object upwards, the work done lifting the object will be greater than if a lesser acceleration is applied across the same distance, but examples in my book don't seem to take this into account. (I am assuming that F = ma, and of course the mass stays constant.) Date: 06/21/2005 at 21:53:08 From: Doctor Rick Subject: Re: Calculus/physics definition of work and lifting Hi, Paul. That's an interesting question. You're dealing with frictionless scenarios, in which the work required to move an object depends only on the difference between the potential energies of the initial and final configurations. In particular, it does not depend on *how*, or *how fast*, you get from the initial to the final configuration. The easiest way to figure the work, or energy difference, is to suppose an infinitely slow movement. This means that the force required to lift the object is equal to the weight of the object: it is not necessary to accelerate it. (A momentary, infinitesimal extra force kicks it into slow movement, and it continues at that slow velocity until the final configuration is reached.) I'm sure you'd like some confirmation that my assertions are meaningful! Can the work done really be independent of how fast the object moves? Well, let's consider your 500 lb beam being moved up 30 feet. Instead of exerting a force just equal to the weight of the beam, I'll apply twice that force, thus accelerating the beam upward at 32 feet/sec^2. It moves upward 15 feet in a time t given by t = sqrt(2s/a) = sqrt(2(15 feet)/(32 feet/sec^2)) = 0.9682 sec At that time it is moving upward at a speed v = at = (32 ft/sec^2)(0.9682 sec) = 30.984 ft/sec Then I remove the upward force entirely, so the beam accelerates downward at 32 ft/sec^2. In another 0.982 seconds, it has gone up another 15 feet and is stationary. Then I can reapply an upward force equal to the beam's weight (that is, I can slide a platform under the beam so it can sit there). I have reached the final configuration. Now, what work have I done? I applied a force of twice the weight, or 1000 pounds, while moving the beam up 15 feet. That's a work of (1000 lb)(15 ft) = 15,000 ft-lb. Then I applied no force, and therefore did no work, while the beam rose on its own the remaining 15 feet. Thus I did no more work, and the total work done was 15,000 ft-lb. Does that look familiar? - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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