Investigating Frictional Forces and Spring Extension

[PhanthomJay]

Yeah.. you're right... but that was just a typo in the previous post, even when I do (m2g-f)/k... I still get the half of the correct answer. When I do it applying energy and work, I get the correct answer... But still... why can't I use Newton's laws to get the correct answer. It makes no sense? Why is it half the answer?

Using Newton's laws the way you have done, assumes that the hanging mass is slowly lowered to its new at rest equilibrium position, with no acceleration. And you calculate the unknowns based on this at rest position.The work energy method, however, assumes that the mass is released suddenly, in which case it will accelerate to a max speed then decelerate to 0 speed at the maximum extension, which is twice the 'equilibrium' extension. These are 2 separate assumptions. Looking again at the simple case of a 2kg mass hanging from a spring of spring constant k =60N/m, and using g=10m/s/s, then for the case where the mass is slowly released, using Newton, F_s = mg = 20N, and since x=F_s/k, x = 1/3m. But for the case where the mass is quickly released from rest, then using energy methods,
1/2kx^2 = mgx, solve x = 2mg/k = 40/60 =2/3m, which is the maximum extension of the spring, and twice the equilibrium extension.

 

[yus310]

Using Newton's laws the way you have done, assumes that the hanging mass is slowly lowered to its new at rest equilibrium position, with no acceleration. And you calculate the unknowns based on this at rest position.The work energy method, however, assumes that the mass is released suddenly, in which case it will accelerate to a max speed then decelerate to 0 speed at the maximum extension, which is twice the 'equilibrium' extension. These are 2 separate assumptions. Looking again at the simple case of a 2kg mass hanging from a spring of spring constant k =60N/m, and using g=10m/s/s, then for the case where the mass is slowly released, using Newton, F_s = mg = 20N, and since x=F_s/k, x = 1/3m. But for the case where the mass is quickly released from rest, then using energy methods,
1/2kx^2 = mgx, solve x = 2mg/k = 40/60 =2/3m, which is the maximum extension of the spring, and twice the equilibrium extension.

... 2/3 is not the answer. But I get what you are trying to explain. I think I have to relate work & energy.

Emech=-Etherm (Energy Dissipated by friction)

Or
Wnet=Wfriction

Ef-Ei=-f*x
(Uf+Ufsp+Kf)-(U+Uisp+Ki)=-f*x
(Uf +Usp+0)-(0+0+0)
1/2kx^2+m2g(-x)+0=-f*x...

... x(1/2kx-m2g+f)=0...

x=0... @ equilibrium position...

or 1/2kx-m2g+f=0...
there is where the half is...

And I get 0.409 m (correct), which is unlike the .204 m, i got from the Netwon's Laws...

I just want to thank "e)hn..." and you, "PhantomJay", for helping me out with this problem and sticking with me. Thanks.

 

[PhanthomJay]

... 2/3 is not the answer. But I get what you are trying to explain. I think I have to relate work & energy.

Emech=-Etherm (Energy Dissipated by friction)

Or
Wnet=Wfriction

Ef-Ei=-f*x
(Uf+Ufsp+Kf)-(U+Uisp+Ki)=-f*x
(Uf +Usp+0)-(0+0+0)
1/2kx^2+m2g(-x)+0=-f*x...

... x(1/2kx-m2g+f)=0...

x=0... @ equilibrium position...

or 1/2kx-m2g+f=0...
there is where the half is...

And I get 0.409 m (correct), which is unlike the .204 m, i got from the Netwon's Laws...

I just want to thank "e)hn..." and you, "PhantomJay", for helping me out with this problem and sticking with me. Thanks.

Yes, the 2/3 and 1/3 numbers I used were just for an example using a different problem. Your numbers are correct, nice work! This problem required a bit of thought for sure.

 

[e(ho0n3]

Using Newton's laws the way you have done, assumes that the hanging mass is slowly lowered to its new at rest equilibrium position, with no acceleration. And you calculate the unknowns based on this at rest position.The work energy method, however, assumes that the mass is released suddenly, in which case it will accelerate to a max speed then decelerate to 0 speed at the maximum extension, which is twice the 'equilibrium' extension. These are 2 separate assumptions.

That is entirely correct. To reiterate: the x you calculated using Newton's laws is the x at the springs new equilibrium position. The x calculated using energy methods is the maximum amplitude (extension) of the spring when the masses are released. To find this latter x using Newton's laws would require fiddling with differential equations I think.