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> It was suggested that Force = mass X velocity.  Also, that Power = velocity
> squared.  (or is it mass X velocity X velocity?)
>
> First, is this stated properly by me.
>
> Second,  can these notions be expressed in a way that my uneducated brain can
> grasp?

Some basic physics definitions:

Momentum is defined as Mass x Velocity.  It has magnitude,
and also direction (since Velocity has direction).

Force is defined as Mass x Acceleration.  It represents how hard
you are pushing (or pulling) on something, and can be measured
with a spring.  It has magnitude and direction.  Note that
Acceleration is (change in Velocity)/Time.  Thus, Force
represents the rate of change of Momentum.

Energy is defined as Mass x (Velocity squared).  It has
magnitude, but no direction.  Note that Energy could also
be defined as Momentum * Velocity -- it's the same thing.

Power is the rate of change of Energy.  So it can be defined
as Momentum * (change in Velocity)/Time, and if you work
it out, this is equivalent to Velocity * Force.

Now, I've been a bit sloppy with the notation "*" here.
It should actually represent a "vector dot" product.
When you multiply two vectors (items with magnitude and
direction), the result depends on the directions, as well
as on the magnitudes.  With a "vector dot" product, the
result of A*B is mag(A)*mag(B)*cos(theta) where theta
represents the angle between the vectors.  So when multiplying
a vector by itself (as when squaring velocity to get energy),
the result is simply the square of the magnitude, since
cos(0)=1.  When multiplying two vectors that point in exact
opposite directions, the result is the negative of the
product of their magnitudes.  And another important special
case occurs when the two vectors are at right angles: the
vector dot product is zero, regardless of the magnitudes,
since cos(90d)=0.

Imagine a ball on the end of a rope, tied to a tall pole.
Stretch out the rope, and give the ball a push, so it rotates
around the pole.  The rope is exerting a force on the ball,
as it is constantly changing (the direction of) the ball's
velocity vector.  But no power is expended, since the direction
of the force is always perpendicular to the direction of the
velocity vector.

Hope this forms a useful starting point...
        -- Bill

   

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