At a given instant in time, a 6 kg. rock is observed to be falling with an acceleration of 5.0 m/s. What is the magnitude of the force of air resistance acting on the rock?
2007-02-01 16:41:54 · 4 answers · asked by damon h 1 in Science & Mathematics ➔ Physics
If not for air resistance, the rock would accelerate at 9.8 m/s/s.
So, 9.8 - 5 = 4.8 m/s/s is being lost to the force pushing in the opposite direction.
F = 6kg * 4.8 m/s/s or 28.8 Newtons.
2007-02-01 16:58:09 · answer #1 · answered by tbolling2 4 · 0⤊ 0⤋
As you know if there was no air resistance then everything should fall at the speed of gravity 9.81m/s. So if the rock is only falling at 5m/s then the air resistance must be (9.81-5) 3.81m/s. But as we are not finding the acceleration and you need the force then all you have to do is multiply it, as F=ma=(6x3.81).
I hope that this answers your question
2007-02-02 00:53:29 · answer #2 · answered by David 1 · 0⤊ 0⤋
sum of the forces equal mass times acceleration
let Far = force of air resistance (acting upward)
assume the only other force is the force of gravity
Far - mg = ma
Far = mg + ma = m(g+a)
m and a are given, and you know g...calculate Far
(in the above equation we have to use a = -5 because we defined Far as positive upward...to make it clearer, write it out in vector form)
2007-02-02 01:03:14 · answer #3 · answered by Anonymous · 0⤊ 0⤋
about 30 newtons
2007-02-02 01:01:53 · answer #4 · answered by anonimous 6 · 0⤊ 0⤋