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Formula for Calculating Work

Date: 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 
Associated Topics:
College Physics
High School Physics/Chemistry

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