12 August 2005

How do forces produce work?

by George E. Hrabovsky, President MAST

A Note to the Reader

You will notice that I ask questions and number them. I have no intention of answering these questions in sequence. The numbering is a bookkeeping device so that I can keep track of things. The point of this column is to investigate what heat is, and I will only go so far afield from that goal. This means that some questions will go unanswered. This is reasonable, and it allows for future projects based on those unanswered questions. Feel free to attempt to answer these questions for yourself.

Where We Have Been

Last time, we explored a crucial mathematical tool. This time we get back to the physics.

The archives and show that your last Theorist column was http://www.sas.org/tcs/weeklyIssues_2005/2005-02-25/mot/

Session 7: How do forces produce work?

Returning to our definition of force we have the famous formula,

F = m a .

We recall that a force is something that accelerates a mass.

Next, we recall our definition of work

W = Δ K,

and we recall that the quantity of work done by forces is equivalent to the change in the kinetic energy of that system throughout the interval where the forces are acting.

What is kinetic energy?

We will not be able to make any progress unless we can define kinetic energy. Let me introduce another free reference [1]. Here we have the kinetic energy of a single particle as,

K = 1/2m v^2 .

So, the kinetic energy depends on the mass and velocity of the particle. We can rewrite this using our newfound knowledge of derivatives,

K = 1/2m (x/t)^2 .

In our discussion of heat we are talking about systems of particles. How do we account for the kinetic energy of all of the particles in a system? We add them all together, one for each particle. To do this I will introduce a short-hand way of writing sums,

a_1 + a_2 + a_3 + ... + a_n = Underoverscript[∑, i = 1, arg3] a_i .

In other words, the sigma symbol represents the sum of the n terms beginning with the first term. For kinetic energy we know that if we have n particles then the kinetic energy must be

K = Underoverscript[∑, i = 1, arg3] 1/2m_iv_i^2

      = 1/2Underoverscript[∑, i = 1, arg3] m_iv_i^2

      = 1/2Underoverscript[∑, i = 1, arg3] m_i(x_i/t)^2 .

If we examine the formula for force,

F = m a,

and the formula for kinetic energy,

K = 1/2m v,

we note that mass is common to each. If we were to solve the formula for force in terms of mass,

m = F/a

and substitute that into our expression for kinetic energy,

K = 1/2F/a v,

or,

K a = 1/2F v .

We can then apply our knowledge of calculus again to get,

K v/t = 1/2F x/t .

Or,

K v = 1/2F x .

This is called a differential equation, and learning how to solve these will be at the heart of our ability to actually do physics.

    50.  What is a differential equation?

Book Review: Physics for Minority Students

Edray Herber Goins, "California Institute of Technology Minority Student Education Freshman Summer Institute 2003 - Physics" (August 2003).

This is a set of physics lectures that have been written down. While they do not spend a lot of time explaining things, they do spend a lot of time working through examples of how to solve problems. That is the key to learning to do theoretical physics. Following along, and explaining each step to yourself, is very useful.

References

[1] Edray Herber Goins, "California Institute of Technology Minority Student Education Freshman Summer Institute 2003 - Physics" (August 2003)


Created by Mathematica  (February 28, 2005)

   
Copyright 2005 by Society for Amateur Scientists