TPR Full Length 5 Help!

[AKMCAT]

If anyone could please help I would really appreciate it. The correct answer is (B); however, I'm not conceptually understanding why.

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[cjcarter]

When you see springs, think Hooke's Law: F= -k Δx

In springs, the force F is the tension in the spring so F=mg. Set mg=-k Δx and solve for Δx. So you get B) mg/k

 

[AKMCAT]

When you see springs, think Hooke's Law: F= -k Δx

In springs, the force F is the tension in the spring so F=mg. Set mg=-k Δx and solve for Δx. So you get B) mg/k

But it says the distance from the top of the box to the bottom of the spring at equilibrium : Shouldn't delta x be zero at equilibrium?

 

[cjcarter]

But it says the distance from the top of the box to the bottom of the spring at equilibrium : Shouldn't delta x be zero at equilibrium?

Yeah I thought that too, this questions sucks lol. But I think since there is a mass attached to the spring, it is still displacing the spring from equilibrium compared to a spring with no mass attached. If you look back at the picture, you can see the distances from the top of the box to the bottom of the spring vary from when it is fully compressed and fully stretched. So when the spring isn't oscillating anymore, it will still be stretched bc of the mass, and I think they want that distance. What is their explanation? I keep staring at this and wondering if I'm thinking about it correctly.

 

[labqi]

Yes, this is something I also found confusing. Is k dependent on Mass? Looking at TPRH passage 44 and my rationale...deltaX is the change in length at equilibrium. So if it is naturally 40 cm long, and changes to 65cm, the deltaX is 15cm. Only at 40cm is the equilibrium 0. I hope that makes sense. The questions aren't clear IMO.

 

[cjcarter]

Yes, this is something I also found confusing. Is k dependent on Mass? Looking at TPRH passage 44 and my rationale...deltaX is the change in length at equilibrium. So if it is naturally 40 cm long, and changes to 65cm, the deltaX is 15cm. Only at 40cm is the equilibrium 0. I hope that makes sense. The questions aren't clear IMO.

'k' is a constant that is characteristic of the spring, it's stiffness. Not dependent on mass.

 

[inasensegone]

If anyone could please help I would really appreciate it. The correct answer is (B); however, I'm not conceptually understanding why.

No offense, but you might be better off putting this in the Q&A sub-forum in the future. Also, I agree that this is poorly worded because the top of the mass and the bottom of the spring should not have a changing length somehow. What the question seems to be attempting, is to ask you to calculate delta x in a convoluted way. I think in trying to make it convoluted though, they dropped the ball and the question doesn't really make sense.

Yes, this is something I also found confusing. Is k dependent on Mass? Looking at TPRH passage 44 and my rationale...deltaX is the change in length at equilibrium. So if it is naturally 40 cm long, and changes to 65cm, the deltaX is 15cm. Only at 40cm is the equilibrium 0. I hope that makes sense. The questions aren't clear IMO.

Even if the question wasn't messed up, and it successfully asked about the change in length at equilibrium, this is still possible. When the spring has a mass suspended from it and displacing it from the equilibrium position of the spring, it is at a new equilibrium (delta x away from the old euilibrium). Thus, "equilibrium" is still valid.

 

[labqi]

Yes. So to be clear equilibrium is whatever it is from the natural length?

 

[AKMCAT]

Yes. So to be clear equilibrium is whatever it is from the natural length?

Yes I think thats what the question is really getting at. So it all depends on what you define as "equilibrium" , here I guess we are defining it as the spring without the mass and considering delta x from the length extended after adding the mass. Correct me if I'm wrong.