Natural Science Forum / Physics / Research / March 2008

In article
<4a7043c1-b239-4e6b-9842-82117093b654@s8g2000prg.googlegroups.com>,
Efthimios <eangelopoulos@gmail.com> writes:

I think your question is based on a misunderstanding.  While
mathematically correct, think of it as  kgr * m/sec/sec.  The sec^2
denotes an acceleration.  Speed is m/s, and acceleration is change in
speed with time, or (m/s)/s.  One can write this as m/sec^2 which will
not lead to wrong results, but one needs to think about what it means.

There are units like J/s/Hz.  They are usually written that way, and it
is clear that it is one Joule per second per 1-Hz bandwidth. A Joule/s
is a watt, so one could also write W/Hz, which is fine.  However, a
Joule is kg*m*m/sec/sec (a similar comment about m^2 applies similar to
that about s^2 above), and 1 Hz is 1/s, so one could write kg*m*m/sec.
Mathematically equivalent, but it disguises what is meant and why one
needs such a unit in the first place.  Of course, directly it is
difficult to think of what kg*m*m/sec could possibly mean.

Efthimios schrieb:

Depends on your aspect what force is doing for you.

Force is thought conventionally beeing equivalent to acceleration*mass.

Cumulated acceleration over time is velocity

v=int dt K/m.

Velocity in your model is the same as that with N/2. The same is true
for the momentum change.

If instead you are looking for the energy transfer growing in time with
t^2 for a constant applied force you get of course a mean square. If the
force is applied for the half time only you have 1/4 of kinetic energy.

Again another mean you get for the travel distance, because then you
have a mix of phases with quadratic and linear increase.

There existist a test I like to apply to physicists, mathematicians and
economists with catastrophic results:

If at your first job your new boss offers you the following alternatives

1: 62000$ per annum + 2000$ salary increase every 2 years
2: 62000$ per annum +  250$ salary increase every half year

what is you choose?

I confess that I needed about 2 weeks to see the analogy myself. Time
matters. Never met anybody to see the solution at first glance.

You said half the TIME, so I can show you need twice the force. Use
impulse and momentum.
    Fdt = MdV = dV    (assuming 1kg.)
Ex1: apply 1 N for 4 seconds and dV reaches 4 m/s.
Ex2: apply 2 N, 0 N, 2 N, 0 N each 1 second, and you get dV of 2 m/s
from each "on" cycle resulting in
    2 + 0 + 2 + 0 = 4 m/s.
You need twice the constant force.
Otherwise, mention another constraint to clarify the question. Your
guess of 1/4th the force indicates there may be more to this.
John Polasek