Measure of spring in system with roller

[Aireve]

Homework Statement

m1= 0,05 kg, m2= 0,03 kg, mass of spring is omitted. We look at movement after spring stopped vibrate. How long does the spring measure during the movement if not stretched spring measure 0,1 meter and if we use force 0,1N elongation measures 0,02 meter.
(sorry, English isn't my main language, I've never learn scientific words so I'm not quite sure if my translation is right)

I know the answer is 0,174 meter.

Homework Equations

The Attempt at a Solution

m1*a = m1*g - Ft (Ft is thread tension)
m2*a = Ft + Fs - m2*g

0,1 = k*0,02
k=5

Don't know what next and if it's even right.

 

[Jamison Lahman]

That type of device is called the Atwood machine. See if this helps: https://en.wikipedia.org/wiki/Atwood_machine

 

[Aireve]

Hym, I can solve problems when there is no spring. That spring is real problem since I don't know how to mark forces at that device. Is Fs=Fg? My friend tried to solve that without Fs but I'm not sure...

 

[Jamison Lahman]

Is Fs=Fg? My friend tried to solve that without Fs but I'm not sure...

Why would Fs be Fg?
Is the spring applying force in either direction? what does Newton's third law tell us?

 

[Aireve]

Why would Fs be Fg?
Is the spring applying force in either direction? what does Newton's third law tell us?

I'm not sure but aren't second mass applying force? Can't Fg be force of action and Fs force of reaction?

 

[Jamison Lahman]

I'm not sure but aren't second mass applying force? Can't Fg be force of action and Fs force of reaction?

Is there any effect from mass 1?

 

[Aireve]

Is there any effect from mass 1?

Well, because of mass 1 that system is moving to the left.

 

[Jamison Lahman]

Well, because of mass 1 that system is moving to the left.

I think you have the right idea. Translation may be a little off. Because of mass 1, the force on mass 2 does not equal the the force of gravity on mass 2.
With that being said, because the spring is without mass it does not contribute to the acceleration of mass 1. The tension in the string, F_t , is therefore also the same force acting on the spring, F_s .

 

[Aireve]

I think you have the right idea. Translation may be a little off. Because of mass 1, the force on mass 2 does not equal the the force of gravity on mass 2.
With that being said, because the spring is without mass it does not contribute to the acceleration of mass 1. The tension in the string, F_t , is therefore also the same force acting on the spring, F_s .

So...
0,1 = k*0,02
k=5

m1*a = m1*g - Ft
m2*a = Ft - m2*g

a(m1 + m2) = g(m1 - m2)
a = 0,2/0,08 = 2,5 m/s^2

Ft = m2(a + g) = 0,03*12,5 = 0,375 = Fs

0,375 = 5*x
x = 0,075 m

x+0,1 = 0,175 m = 17,5 cm

Is it correct?

 

[Jamison Lahman]

So...
x+0,1 = 0,175 m = 17,5 cm

Is it correct?

Looks like it

 

[Aireve]

Looks like it

Ok, thank you!
But can I have one more question? How should I draw those forces at that picture? Should I draw Fs and Ft or only one of them? They should "touch" mass 2 or spring?

 

[Jamison Lahman]

Ok, thank you!
But can I have one more question? How should I draw those forces at that picture? Should I draw Fs and Ft or only one of them? They should "touch" mass 2 or spring?

You could explicitly show all of the forces (so that the spring looks like the green to the side):

But I think that would be messy and overly complicated. I personally would only draw forces on objects with mass.

 

[Aireve]

You could explicitly show all of the forces (so that the spring looks like the green to the side):

But I think that would be messy and overly complicated. I personally would only draw forces on objects with mass.

Right, I don't think I have to do that much.
To be 100% sure, it's ok?

 

[Jamison Lahman]

Looks fine to me

 

[Aireve]

Ok, thank you. :)