Integration and the Concept of WorkDate: 3/19/96 at 20:5:42 From: J. DeMeo Subject: (no subject) In calculus class we are studying work. If asked to find the work done by a varying force F(x) = 4x^3 from -4 to +4, using integration the work done equals 0. Mathematically I understand why that is, yet how can it be physically that no work is done when applying this force over a distance of 8 units? The calculus book says the answer really is zero, but how do you convince someone that this makes sense? Is there a way to make an analogy in the real (physical) world? Thanks Date: 3/21/96 at 13:17:59 From: Doctor Aaron Subject: Re: (no subject) Hi, The concept of work in physics is a little different from everyday usage, so it takes a little getting used to. One way to look at it is to break the integral up into 2 regions: 4 0 4 :: :: :: W= -: F = -: F + -: F. If you evaluate both of these integrals, you :: :: :: will get 256 + -256 = 0. -4 -4 0 The physical interpretation of this is that, while the object is on the left (negative) side of the x-axis, the force is negative, so the object is being pushed in a negative direction. By a physicist's definition, work is being done (256 units worth of it). However, when the object is on the right (positive) side of the axis, it is being pulled away from the origin so negative work is being done (again, 256 units worth). This idea of negative work doesn't make sense if you think of work as stuff that makes you tired, but in physics it is very useful. There is one other thing in my explanation that should confuse you. I said that the object gets pushed away while it is on the negative side of the axis. You might thing that it should never get to the positive side. Now think of throwing the object from -4 on the x-axis, the force (4x^3) will be negative until it gets to the origin. This force will slow down the object a great deal, but once it gets to the positive side of the axis, the forces will push the object in the positive direction. It turns out, because of the anti-symmetry about the origin of the way that the force acts, that once the object gets to 4 on the x-axis, it will have exactly the same speed as it did at -4, when you threw it (provided you threw it fast enough that the object made it to the origin without turning around). If it is helpful, you can think of a configuration of planets that produce a gravitational force field, or a configuration of charged objects that produce an electro-magnetic force field, such that the force is 4x^3 on the x-axis between -4 and 4. If a meteor or charged object flew through this force field, it would have the same velocity at -4 and +4 on the x-axis. I hope that this is helpful, if you need further clarification, don't hesitate to write back. -Doctor Aaron, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/