# Where is the fault in the mechanical energy conservation in a spring block system here?

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 ?

• The two situations are different, the total energy of the first is not related to the second situation. You added energy to the system in the second. Commented Jan 16, 2022 at 8:37
• How is the spring elongated? Commented Jan 16, 2022 at 8:37
• Does this answer your question? Different Spring constant when calculating different ways? Commented Jan 16, 2022 at 9:23
• Hello! It is preferable to use MathJax (LaTeX) to display formulas. You can find a tutorial at MathJax basic tutorial and quick reference. Please edit your question accordingly. Thanks! Commented Jan 16, 2022 at 10:50
• Suppose I added energy externally through a conservative force . Then , mechanical energy must be conserved even when external forces are applied because only conservative forces are involved Commented Jan 19, 2022 at 5:36

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