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aishax
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A spring is suspended from a celing and a 0.256-kg mass is attatched to it and pulled down to stretch the spring by 0.182 m. The mass is released and travels through the equilibrium position with a speed of 0.746 m/s. Calculate the force constant on the spring.
I'm not really sure how to tackle this question.
What I ended up trying was looking at the two points being addressed (the point where the spring is stretched 0.182 meters downwards, and the equilibrium).
Using the conservation of energy, I put the equation:
E total1 = E total2
(Where E total1 is at the point where the spring is stretched, and E total2 is the point at which the spring passes the equilibrium)
I then put:
Ep + Ek + Ee = Ep + Ek + Ee
I canceled out Ep ( because h = 0), and Ek (because v = 0) on the left side of the equation. I canceled out Ee on the right side of the equation because x = 0. The equation became:
Ee = Ep +Ek
So far, am I on the right track, or is what I did completely wrong? Could this question be done a simplier way using the formula: Fs = -kx ?
Help is very much appreciated :)
I'm not really sure how to tackle this question.
What I ended up trying was looking at the two points being addressed (the point where the spring is stretched 0.182 meters downwards, and the equilibrium).
Using the conservation of energy, I put the equation:
E total1 = E total2
(Where E total1 is at the point where the spring is stretched, and E total2 is the point at which the spring passes the equilibrium)
I then put:
Ep + Ek + Ee = Ep + Ek + Ee
I canceled out Ep ( because h = 0), and Ek (because v = 0) on the left side of the equation. I canceled out Ee on the right side of the equation because x = 0. The equation became:
Ee = Ep +Ek
So far, am I on the right track, or is what I did completely wrong? Could this question be done a simplier way using the formula: Fs = -kx ?
Help is very much appreciated :)
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