0

Consider a spring block system . The surface on which the block moves is frictionless and the air resistance is nil . Suppose the block is at the mean position in it's natural length . By hookes law , ( PE ) = 0 because x = 0 . Also , the ( KE ) = 0 because block is at rest . Hence , total mechanical energy = KE + PE = 0 . Now , suppose it is elongated to maximum elongation . Then , at the point of maximum elongation ; KE = 0 and PE = k.a² / 2 , where a is the maximum displacement of elongtion . Hence , at this point , Mechanical energy = KE + PE = 0 + k.a²/2 = k.a²/2 . It is evident that initial mechanical energy ( = 0 ) is not equal to final mechanical energy ( ka²/2 ) . But how ?

Qmechanic
  • 201,751
Abbas
  • 239

1 Answers1

3

Energy is added to the system by the force that pulled the mass to position $a$.

Using conservation of energy to make predictions about a system requires the concepts of “closed” and “open” systems.

cms
  • 3,998
  • No, just look at system , not external – Abbas Jan 16 '22 at 07:51
  • No , just look at system . There are only conservative forces . So mechanical energy should be conserved – Abbas Jan 16 '22 at 07:51
  • @Abbas In your post, you said, "Also , the ( KE ) = 0 because block is at rest." If the block starts at rest with the spring at natural equilibrium length and if your system is isolated with no external forces, then the spring is never elongated because the block would continue to be at rest for all time. By Newton's first law, if something is at rest, it continues to be at rest unless there are external forces. – Maximal Ideal Jan 16 '22 at 08:41
  • Assume that the external forces are conservative . Then ? – Abbas Jan 19 '22 at 05:37
  • @Abbas then there would be a potential energy associated with that force. – cms Jan 20 '22 at 00:22
  • No . Let me explain with an example . – Abbas Jan 20 '22 at 03:10
  • Let there be spring block system in space ( gravity free ) in it's natural length and let KE of system be zero as viewed from a frame . Now , initial mechanical energy = 0 because KE = PE = 0 . Now , bring a planet near it so that It's gravity act on a particle . Now , since only conservative forces are involved , hence , the final mechanical energy must be zero as viewed from the reference frame . But , as you said , it had gained some P.E. , hence Conservation of mechanical energy is failed . – Abbas Jan 20 '22 at 03:14
  • @Abbas Gravitational PE is always negative. So in your example, you start with 0 KE and 0 PE and end with positive KE and negative PE. So the net energy is still zero. – cms Jan 20 '22 at 20:38
  • No , there is no positive KE – Abbas Jan 21 '22 at 17:50
  • @Abbas how? There then is an external force acting on the object to prevent it from moving under the influence of gravity and you have an “open” system. Or your spring stretches and you gain positive spring PE. At any rate, your confusion comes from not using the concepts of closed and open systems correctly. Those concepts must be used for conservation of energy. – cms Jan 21 '22 at 18:39
  • the force which prevents it from moving under influence of gravity is spring force which is not external – Abbas Jan 22 '22 at 10:28