equation of potential energy in the spring in equlibrium position will be
Answers
Answered by
1
Suppose that an object AA with mass M=8 kg at 10 mt above the floor falls from rest tied to an 2 mt elastic rope, which does not exert any force since AA falls 2 (and I will model it as a spring with elastic constant kk). I know it will bounce at a minimum height - doesn't matter where it is, but let it be y1y1-, and at that moment the object decrease its mass in 3 kg, so that new mass is m=5 kg.
I want to know the maximum height it will reach after bouncing, so I set up conservation of energy (assuming there is no friction due to the air). Now, which is the elongation of the spring? I mean, we measure it respect to the equilibrium position. Initially, but it changes when mass changes (it is a a higher position when AAweighs less). Although, if elongation is higher after losing mass, I think it implies system has more potential than before but no force has done work to change it. It seems like system has gained energy, which looks absurd.
So, what elongation should I use when I set up the eq:
Mgy1+12k(yeq−y1)2=mgy2+12k(yeq^−y2)2Mgy1+12k(yeq−y1)2=mgy2+12k(yeq^−y2)2
Here some images to illustrate this situation.


Where y2y2 will be the maximum height after bouncing.
Remember yeqyeq is the equilibrium position when AA has mass M, and yeq^yeq^ is the same when AA has mass m.
I hope you could help me to see it more clear. Please, correct me if I'm wrong.
Plz mark the answer as brainly answer
I want to know the maximum height it will reach after bouncing, so I set up conservation of energy (assuming there is no friction due to the air). Now, which is the elongation of the spring? I mean, we measure it respect to the equilibrium position. Initially, but it changes when mass changes (it is a a higher position when AAweighs less). Although, if elongation is higher after losing mass, I think it implies system has more potential than before but no force has done work to change it. It seems like system has gained energy, which looks absurd.
So, what elongation should I use when I set up the eq:
Mgy1+12k(yeq−y1)2=mgy2+12k(yeq^−y2)2Mgy1+12k(yeq−y1)2=mgy2+12k(yeq^−y2)2
Here some images to illustrate this situation.


Where y2y2 will be the maximum height after bouncing.
Remember yeqyeq is the equilibrium position when AA has mass M, and yeq^yeq^ is the same when AA has mass m.
I hope you could help me to see it more clear. Please, correct me if I'm wrong.
Plz mark the answer as brainly answer
Answered by
1
EQUILLIBRIUM means state where all forces are balanced so if you attached a weight to a spring and relesed it you see it start to oscillate but some times later it stops and be stable so =>P.E=maximum elongation of spring due to weight P.E=1kx^2/2
Similar questions