Gary Novak

Energy Home

6. Falling Objects

The numbers on this page are used for proof
on the rocket page.

About three hundred years ago, Leibniz stated a premise upon which the underlying assumptions about kinetic energy are base. He said in effect that if a four kilogram object were dropped one meter, it would produce the same result as a one kilogram object dropped four meters. He then attempted to show mathematically that such a result conserves mv² but not mv.

Leibniz rearranged the proportions of force and distance while maintaining the same totals. He thereby "conserved"; the force-distance combination. Conservation in physics means that the same totals are found after an event as before, even when the component elements are rearranged.

Nowdays, there are said to be two conservation laws—the conservation of energy and the conservation of momentum. The historical development of the definition of energy attempted to prove that momentum is not conserved, which contradicts modern concepts.

The purpose of this page is to compare the Leibniz analysis with the alternative which Descartes might have stated had he been around in Leibniz's time.

What Descartes might have said is that a four kilogram object dropped for one second will produce the same result as a one kilogram object dropped for four seconds. The result would have shown the conservation of momentum and produced a correct definition of kinetic energy, where momentum and kinetic energy are the same things.

First, the basic principles will be described.

The Basics.

Let's first consider why force times distance is proportional to ½mv². Such a relationship only exists while a force is accelerating a mass. The counterforce is then inertia, which is sometimes referred to as unbalanced force.

To analyze the dynamics of motion, these equations are used: (calculus is being avoided because it masks the logic.)

 F = ma v = s/t = at s = ½vt = ½at²

Where:

 F = force v = velocity m = mass t = time a = acceleration s = distance (or seconds)

Consider as an example a 5kg mass accelerated at 10m/s/s for 4 seconds.

F = ma = 5(10) = 50 newtons

v = at = 10(4) = 40 meters per second

s = ½vt = ½(40)(4) = 80 meters

(The ½ averages the uniformly changing velocity.)

Fs = 50(80) = 4000 newton-meters

½mv² = ½(5)(40)² = 4000 joules

The quantities of Fs and ½mv² are the same. The reason why is established by factoring (or calculus):

Fs = (ma)(½vt) = ½mv(at) = ½mvv = ½mv²

Leibniz's original premise was that conserving Fs would also conserve mv². He did not make a direct mathematical equality, because without the ½ they are not equalities.

With the ½ being included, there is no question of Fs being proportional to ½mv². The real question is why conserve Fs. In fact, why should Fs even be measured or analyzed? There is no logic to analyzing Fs, because the force does not move through any distance relative to the mass it acts upon. The distance factor relates the force to the starting point, while the force has a different relationship to the staring point than to the mass it is acting upon. In other words, the combination of force and distance is a meaningless quantity which is factored into another meaningless quantity, ½mv².

The reason why ½mv² is conserved during elastic collisions is because the force changes at the same time velocity changes producing a compounded effect upon the change in velocity resulting in both factors being conserved. But such a situation does not exist during inelastic collisions, where the force may be constant, and the quantities mv and ½mv² cannot both be conserved.

Leibniz stated that his analysis did not conserve mv. So let's consider an example which demonstrates why mv and ½mv² cannot both be conserved for falling objects.

Force-Distance Combination.

The metric system will be used comparing a 4kg mass (m1) dropped 1 meter with a 1kg mass (m2) dropped 4 meters. The force acting upon each mass is determined as F = mg. The quantity g is the acceleration of gravity, which is 9.81m/s/s; and the resulting force is in newtons. (g is 9.80665; but rounding is OK here, when always the same, because the test is valid at any consistent gravity level.)

For time dropping:

t =
 √ 2s/g
(from s = ½at²)

For velocity down: v = gt

For m1 (4kg):

F = 4(9.81) = 39.2

t =
 √ 2(1)/9.81
= 0.452

v = 9.81(0.452) = 4.43  (for proof)

mv = 4(4.43) = 17.7

½mv² = ½(4)(4.43)² = 39.2

For m2 (1kg):

F = 1(9.81) = 9.81

t =
 √ 2(4)/9.81
= 0.903

v = 9.81(0.903) = 8.86  (for proof)

mv = 1(8.86) = 8.86

½mv² = ½(1)(8.86)² = 39.2

So when the force-distance combination is conserved, the ½mv² is the same for both masses (39.2), while the mv is not (17.7 vs 8.86).

Force-Time Combination.

If however, the force-time combination were conserved rather than the force distance combination, the opposite would be true. We will compare a 4kg object dropped for 1 second to a 1kg object dropped for 4 seconds. The forces are the same as before: 39.2N for m1 and 9.81N for m2.

For m1 (4kg):

v = 9.81(1) = 9.81  (for proof)

mv = 4(9.81) = 39.2

½mv² = ½(4)(9.81)² = 192  (compare)

For m2 (1kg):

v = 9.81(4) = 39.2  (for proof)

mv = 1(39.2) = 39.2

½mv² = ½(1)(39.2)² = 770  (compare)

The mv is the same for each mass, while the ½mv² is not. The reason why mv is conserved is because Ft was conserved; and Ft = mv. In other words, the amount of time that a force acts upon a mass determines its momentum afterwards. Demonstrating that Ft = mv is simply to factor equations.

Ft = (ma)t = m(v/t)t = mv

The Meaning.

So the question is, why conserve the force-distance (Fs) combination rather than the force-time (Ft) combination. Perhaps because Fs is more apparent and easier to visualize. A mass at a height of 1 meter or 4 meters is easy to visualize mentally. But a mass which falls for 1 second or 4 seconds is not so easy to visualize. Time does not create as clear of a mental image as distance does.

A fall for 1 second requires a height of 4.9 meters; and a fall for 4 seconds requires a height of 79 meters. A person does not know that until it is calculated. (The formula is h = ½gt²). So there was probably a prejudice towards conserving Fs rather than Ft.

The reason why mv and ½mv² could not both be conserved for falling objects is because there was no elastic collision. With gravity, the force stays constant rather than varying with the velocity; and therefore, mv and ½mv² each have different dynamics. With a constant force, mv increases linearly, while ½mv² increases exponentially, as demonstrated later with engines. Since they have different dynamics, rearranging one by a certain proportion does not rearrange the other by the same proportion.

There is no simple experiment which can prove whether one of the falling masses will produce the same result as the other. Leibniz assumed that the two masses would produce the same force. However, the force resulting from a collision would depend upon the hardness of the materials. The only link creating a relationship between the two masses which Leibniz described was the appealing symmetry of conserving the force-distance combination.

The problem with conducting an experiment to test the result is that there is no way to get all of the energy of one mass transferred to the other mass. If one mass were to collide with the other mass elastically, the first mass would still have motion. If an inelastic collision occurred, some of the energy would be converted to heat, either as a net reduction in ½mv² or a reduction in momentum relative to the impact point, depending upon definitions.

Applying Newton's Laws.

It should be obvious that kinetic energy can only be transferred or transformed through force, because only force can change the motion of a mass. If the issue can be reduced to the question of what result force will produce, the answer is determined by Newton's laws.

One of Newton's laws is that for every force there must be an equal and opposite force. Forces can only be equal and opposite if they act for equal amounts of time. Therefore, there must always be the same amounts of the force-time combination on both sides of any interaction involving force; and the force-time combination can never be broken where forces interact. And therefore, momentum must be conserved, because it is a product of Ft.

When Newton's laws are applied to the situation described by Leibniz, and one theorizes the transfer of energy from one mass to the other while maintaining equal and opposite force on each, the Fs combination, or ½mv², is not conserved, while Ft, or mv, is.

For example, if the 4kg object falls 1 meter and then is stopped by a force, the force could arbitrarily be any level, such as 100N. Later, a rocket will be used to produce the force. The rate of deceleration of m1 is:

a = F/m = 100/4 = 25m/s/s.

The amount of time for which the force exists is:

t = v/a = 4.43/25 = 0.177s

So Ft = 100(0.177) = 17.7, which is the same as the momentum for m1 shown earlier. This amount of Ft stops the motion of m1. If it is then applied to m2, the acceleration of m2 is:

a = F/m = 100m/s/s

The resulting velocity of m2 is:

v = at = 100(0.177) = 17.7m/s

The time that m2 could travel (decelerate) upward against gravity with that starting velocity is:

t = v/g = 17.7/9.81 = 1.81s

The distance upward that it could go is:

s = ½gt² = ½(9.81)(1.81)² = 16 meters

So the distance (16 meters) is four times greater than Leibniz's premise indicates (4 meters for the 1kg mass). In other words, a 4kg mass dropped from 1 meter would produce the same result as a 1kg mass dropped from 16 meters, when the result is determined through equal and opposite forces on each. This result stems from the fact that equal and opposite forces require conserving the force-time combination rather than the force-distance combination.

Even though a collision could not create such a situation, external sources of energy could, as shown later with rockets.

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