# Applications of Newton's Laws

## Homework Statement

A meteor of mass .25 kg is falling vertically through Earth's atmosphere with an acceleration of 9.2 m/s. In addition to gravity, a vertical retarding force (due to the frictional drag of the atmosphere) acts on the meteor. What is the magnitude of this retarding force?

## Homework Equations

F = ma

but I think, because of "in addition to gravity"

F = ma + mg

## The Attempt at a Solution

Fm-Ffr= ma + mg

Fm = (.25)(9.2) = 2.3 N

-Ffr= ma + mg - Fm

-Ffr = mg

= -2.45 N
the friction force greater than the weight? Doesnt sound right

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Doc Al
Mentor
F = ma

but I think, because of "in addition to gravity"

F = ma + mg
No, you were right the first time:
ƩF = ma

## The Attempt at a Solution

Fm-Ffr= ma + mg
Fix this.

So, Fm-Ffr= ma

-Ffr = ma - Fm

But that equals 0. What other forces are in act here?

Doc Al
Mentor
So, Fm-Ffr= ma

-Ffr = ma - Fm

But that equals 0. What other forces are in act here?
Why do you think it equals 0? (I assume Fm is the object's weight, right?)

Fm is the meteor's weight, yes.

So gravity plays no part in this?

Now I am thinking this
(im switching to Y+ up, i forgot to consider that i was making Y+ down)
ƩF = ma

Ffr - Fm - w = may

Ffr = may + Fm + W

I think this is correct because there is 1 force pulling the meteor back, the Force of friction, and 2 forces acting down, the weight of the meteor(with earths gravity) and the meteor moving with the 9.2 m/s2 acceleration.

Or am I just adding the weight of the meteor twice?
I think I am overthinking this.

Doc Al
Mentor
Fm is the meteor's weight, yes.

So gravity plays no part in this?
What do you think gravity is?

Okay, I'm settling with this.

Ffr - Fm = may

Ffr = may + Fm

Ffr = (.25kg)(9.2m/s2) + (.25kg)(9.8m/s2)

Ffr = 4.75 N

Doc Al
Mentor
Okay, I'm settling with this.

Ffr - Fm = may

Ffr = may + Fm

Ffr = (.25kg)(9.2m/s2) + (.25kg)(9.8m/s2)

Ffr = 4.75 N
Almost. What's the sign of the acceleration?

negative

Ffr = (.25)(-9.2) + (.25)(-9.8)

Ffr = -4.75 N

Thanks a bunch!

Doc Al
Mentor
negative

Ffr = (.25)(-9.2) + (.25)(-9.8)

Ffr = -4.75 N

Thanks a bunch!
The acceleration of the object is -9.2; the weight is just mg, not -mg. (You already included the direction of the weight--downward--in your first equation.)

One more time!

Ffr = (.25)(-9.2) + (.25)(9.8) = .15 N

finally, 5th times the try.

Thank you very much for the help

Doc Al
Mentor
Ffr = (.25)(-9.2) + (.25)(9.8) = .15 N

finally, 5th times the try.

Thank you very much for the help
Good! And you're welcome.

It's always useful to do a 'sanity check' of your answers. In this case, you know the acceleration is just a little bit less than free fall acceleration. That should tell you that the resistance is small compared to the weight.

I 'am curious would the total force be: Fm+Ffr with Fm=(m*-g)+(m*-a)+Ffr. Therefore Ft=-4.6N

Doc Al
Mentor
I 'am curious would the total force be: Fm+Ffr with Fm=(m*-g)+(m*-a)+Ffr. Therefore Ft=-4.6N
No. Only two forces act: gravity and the retarding force. You can find the net force directly using ∑F = ma.