(1) Newton’s Second Law

Momentum is defined as the product of mass *m* and velocity **v**. The mass of a body is a scalar, though not necessarily a constant. Velocity is a vector equal to the time rate of change of location, **v** = d**s**/d*t*.

The time rate of change in momentum is d**p**/d*t* = *m* d** v**/d

*t*+

**v**d

*m*/d

*t*= m

**a**+

**d**

*v**m*/d

*t*by the rules of differential calculus and the definition of acceleration,

**a**.

If mass is constant, then **v** d*m*/d*t* equals zero and the equation reduces to d**p**/d*t* = *m***a**. If we define **F** = d**p**/d*t*, then we get Newton’s famous **F** = *m***a**.

The dual equation is derived similarly:

*Levamentum* is defined as the product of *vass* *n* and lenticity **u**. The *vass* of a body is a scalar, though not necessarily a constant. *Lenticity *is a vector equal to the space rate of change of *chronation*, **u** = d**t**/d*s*.

The space rate of change in levamentum is d**q**/d*s* = *n* d** u**/d

*s*+

**u**d

*n*/d

*s*=

*n*

**b**+

**d**

*u**n*/d

*s*by the rules of differential calculus and the definition of

*relentment*,

**b**.

If *vass* is constant, then **u** d*n*/d*s* equals zero and the equation reduces to d**q**/d*t* = *n***b**. If we define **R** = d**q**/d*s*, then we get the dual of Newton’s second law, **R** = *n***b**.

(2) Work and kinetic energy

Work is defined as a force **F** applied over a displacement **s**: *W* ≡ **F** · **s**, and for a constant force and displacement in the same direction: *W* = *F**s*. From Newton’s second law, *F* = *ma*, so we have *W* = *mas*.

The third law of motion states that *v*² – *v*_{0}² = 2*as*. Substitute this in the above equation to get

*W* = *m* (*v*² – *v*_{0}²)/2 = *mv*²/2 – *mv*_{0}²/2 = ΔE_{k}.

This is a form of the work-energy theorem, which states that the work done on a body is equal to the change in that body’s kinetic energy.

If the initial kinetic energy is zero, then the kinetic energy E_{k} = *mv*²/2.

Dual to this is repose and kinetic lethargy:

Repose is defined as a release **R** applied over a dischronment **t**: *Y* ≡ **R** · **t**, and for a constant release and dischronment in the same direction: *Y* = *R* *t*. From the dual to Newton’s second law, *R* = *nb*, so we have *R* = *nbt*.

The third law of motion states that *u*² – *u*_{0}² = 2*bt*. Substitute this in the above equation to get

*Y* = *n* (*u*² – *u*_{0}²)/2 = *nu*²/2 – *nu*_{0}²/2 = ΔQ_{k}.

This is a form of a repose-lethargy theorem, which states that the repose done applied to a body is equal to the change in that body’s kinetic lethargy.

If the initial kinetic lethargy is zero, then the kinetic lethargy Q_{k} = *nu*²/2.