I can't remember what the term is for the mechanical equivilent of activation energy. IE if I was pushing the proverbial cart up the hill, it would take slightly more force to get it started than to continue. What is this extra force called?
2006-10-30 08:11:00 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The force is needed to overcome inertia is the same regardless if the mass is stationary or is moving with a constant speed. So the force is the same if you want to get an object at rest of mass m moving at a speed of say 10m/s in 10 seconds will be the same as force required to move an object to speed of 20m/s in 10 seconds if it was moving at 10m/s.
F=ma=m(v2-v1)/(t2-t1)
Case 1(stationary)
F=m(10 - 0)/(10)=m (newtons)
Case 2(in prior motion)
F=m(20 -10)/(10)=m (newtons)
2006-10-30 08:15:26 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
I'm not sure if there is a technical term for this extra force.
But its due to the fact that the static coefficient of
friction is almost always highter than the kinetic coefficient
of friction between two surfaces. (I'm sure there are some
weird surfaces that someone has come up with in the lab where
this is not the case!).
So the initial force required to get somethign moving is higher
than the force needed to keep it moving at a constant rate.
2006-10-30 18:11:16 · answer #2 · answered by Jim C 3 · 0⤊ 0⤋