# An Algebraic discussion of Work and Energy

This discussion parallels the post on "Work and Energy", but uses algebra instead of calculus. The discussion is simplified by making simplifying assumptions to arrive at the same conclusions. The discussion involving calculus is more general and not so specialized. Simplicity here is gained by a loss of generality.

## Preliminaries

This discussion builds onto that contained in a previous post  Grokking Galileo's Physics part II. It does so in a way that Galileo could have understood because he did not know calculus, which was developed by Newton and Leibnitz. However, Galileo was a pretty smart guy, and would have been able to learn calculus fast, in which case he would have probably preferred the discussion involving calculus. Here we use a very simplified model for our  discussion.

### Limiting Assumptions

Consider a mass m, which is moving to the right along a straight line, the x-axis, starting from x=0 at t=0 with an initial speed of  uo. A constant force F acts to the right along this line, which produces an acceleration a.

## Derivation

At a constant acceleration, a, and at any time, t > 0, the mass has a speed as a function of time,

u = uo + a t.                               (1)

The average speed of the mass during the time, t, is

<u>=½ (u + uo).                       (2)

The distance travelled from the origin (the position at t=0) is just

x= <u> t.                                    (3a)

Substituting (2) in (3a), we have

x= ½ (u + uo) t ,                      (3b)

which with the further substitution of (1) reduces to

x = uo t + ½ a t2.                     (3c)

The above (3c) is just the equation for the position of a mass moving under constant acceleration to the right along the x-axis after starting from the origin, with an initial velocity of uo.

It is possible to eliminate time from the description of this motion by solving for it using (1),

t = (u - uo )/a ,                       (4)

and plugging that result and (2) back into (3),

x= ½ (u + uo) (u - uo )/a .    (5a)

carrying through the multiplication and solving for u in terms of x, we get the following:

u2 =   uo2 + 2 a x.                  (5b)

At this point, multiplying both sides of (5b) by  ½ m gives the results,

½ m u2 = ½ m uo2 + (m a) x.   (6a)

Identifying the term in parenthesis on the right hand side (RHS) of (6a) as the force, from Newton's Laws of motion, results in,

½ m u2 = ½ m uo2 +F x.           (6b)

By defining the work done by the constant force F as,

W= F x                                (6c)

and kinetic energy as,

KE=½ m u2,                         (6d)

we can further simplify (6b) to

KE =  KEo +W.                       (7)

Note that (7) identifies work as the transfer of energy through the agency of an external force acting to change the state of motion of a mass.

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