0

There is a spring with springs constant $k =200 N/m$. The spring is held vertically and a mass of 2 kg is attached to it. We will assume that $g=10 m/s^2$. The extension of the spring is to be found. The same can be found in two different ways.

  1. kx = mg --> x = 2*10/200 = 0.1 m
  2. The second method is using energy. The 2kg mass would lose GPE (loss of height = x), which would go into the spring as its strain potential energy. mgx = 0.5kx^2 --> x = 0.2m

Why are the two answers are different? I know that the first answer is correct, but why is the second method not correct? In fact, no matter what numbers I take, the second answer is always twice the first one.

Thanks in advance

Golden_Hawk
  • 1,020
user15041899
  • 1

1 Answers1

0

If you hang 2kg mass at a spring it moves beyond the equilibrium point and starts to oscillate around the 0,1m.

trula
  • 6,146
  • Thanks. that is right, but i am talking about the equilibrium position. We gently lower the mass so that there are no oscillations. In this position, the mass has lost height by x and the resulting loss of GPE got transferred to springs. – user15041899 Aug 25 '23 at 13:54
  • If we "gently" lower the mass, we extend a force to it, energy is not conserved. – trula Aug 25 '23 at 14:08
  • When Ithink about this it makes sense now. Thanks – user15041899 Aug 25 '23 at 22:15