A 30.0N stone is dropped from a height of 10.0m and strikes the ground with a velocity of 7.00m/s. What avera

Question

I am doing Physics I. I would need to know the average force of air friction as it acts on it as it falls?

Answer 1

This is not question of Phys. I.

You must to use this:

F = ma -bv, wher b is a coeffient of friction and v is the velocity. You must to know the coefficient b and resolved the differential equation...

-bv is the "force air"

Answer 2

Harsh people! There's nothing wrong with the question, it's saying the stone 'weighs' 30.0N!!
You don't need to know the mass, but it's given by w=mg.

Calculate the acceleration it fell with:
V^2 = Vo^2 + 2*a*y where V= 7.00m/s, Vo=0, y=10.0m. So
a = 7.00^2/10.0 m/s^2 = 4.90 m/s^2

As pointed out by just "JR", that's about half the acceleration due to gravity so the net force that pulled it to the ground must be about half its weight. Actually 4.90/9.81 = .499 is the ratio, so the force of air resistance was 30 - 30*.499 = 15.0 N

Answer 3

Well, you need to know the mass to calculate the amount of force that is being applied at the latter part of the problem I think. From there, you find the difference in the force, (In Newtons) from the stone at the start and the stone at the end of the drop. That tells you the amount of air particles resisting it. K? Lol, btw, I learned this last year in 8th grade, and it was in Physics. So ya, wouldn't be too complex.

Answer 4

Let suppose it is a 3Kg-mass stone, since you express a mass in Newtons (!?)...
Dropped from a height of 10m, its velocity, calculated without taking account of the air friction, would be:
v = sqr(2.e.g) = sqr(2*10*9.81) = sqr(196.2) = 14 m/s
(v=differential velocity, e=height, g=gravity)
Since to stone reaches the ground at only 7m/s, it has been submitted to a force upwards (the air resistance).
Two forces are acting on the stone:
Fg, which is due to gravity, = mass * acceleration = 3 * 9.81
Fr, the air resistance, = mass * "deceleration" = 3 * x
Since the stone reaches the ground at HALF its speed in the vacuum, it has been submitted to HALF acceleration...
Hence the air "resistance" is causing a deceleration of 9.81 / 2 or
5 ms-2.
Bloody resistant air or very low density stone with parachute shape if I may say...

Answer 5

You won't ever consider that in Physics I because it is far too complicated for you at that level.

In particular, it would be a bit advanced for you since you don't seem to know that N (Newtons) are units of force and not mass. A 30.0N stone is as meaningful as a 2 liter long piece of string.

And, in any event, you haven't said what it *is* that you need to calculate.


Doug

Answer 6

weight of stone - drag force due to airresistance = downward force acting on stone

mg - F = ma
(30) - F = (30/9.81) a------ (1)

resultant acceleration a
v^2 = u^2 - 2as v =final velocity u =intial velocity s= height of fall
(7)^2 = 0 - 2(a)(10)
a = 2.45 m /s^2

sub a= 2.45 into ---(1)
30 - F = (30/9.81)(2.45)
drag force/ air friction = 22.6N

~darbi