A masslessHooke\'s Law spring has unstretched length of 1.750 m. Whena 37.5 kg m

Question

A masslessHooke's Law spring has unstretched length of 1.750 m. Whena

37.5 kg mass isplaced on it, and slowly lowered until the mass is at rest,the

spring issqueezed to a length of 1.712 m.

Mg(1.750 m) = mg(1.712 m)+ ½ k(.0380 m)^2

13.965 kg m^2/s^2 = ½ k(.0380 m)^2

-Kx – mg = 0

-Kx = +mg

-k(-.0380 m) = + (37.5kg)*(9.80 m/s^2)

QuestionDetails: Hi, I worked out a solution to this problem, but I do notunderstand why I am wrong.

A masslessHooke's Law spring has unstretched length of 1.750 m. Whena

37.5 kg mass isplaced on it, and slowly lowered until the mass is at rest,the

spring issqueezed to a length of 1.712 m.

Iwant to find the spring constant (k), so I approachedit like this: Initialenergy = mghi Final energy= mghf + 1/2 kx^2

Mg(1.750 m) = mg(1.712 m)+ ½ k(.0380 m)^2

13.965 kg m^2/s^2 = ½ k(.0380 m)^2

k =19342 kg/s^2 However, the "real" solution found k like this:

-Kx – mg = 0

-Kx = +mg

-k(-.0380 m) = + (37.5kg)*(9.80 m/s^2)

K = 9671kg/s^2 Why would my approach to finding k be wrong? Myapproach is wrong because of the 1/2 in front of the k in theequation. But other than that, it seems my approach isright. Please help me.

Explanation / Answer

as there should be external force to lower the mass slowly you should not equate initial energy and final energy.

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