Wednesday, 26 October 2016

newtonian mechanics - Where is the energy lost in a spring?



Thinking about springs, and their extensions, I recently came to a confusion which I hope this wonderful community can help me solve.


the figures The question is this. When the block is initially attached to the spring, the spring has some extension $x_0$. Now the spring gets extended to some extension $x=\frac{mg}k$ by an external force maintaining equilibrium at all the points such that $KE=0$ at the bottom.


As my reference is the line shown in the figure, the initial potential energy $U$ is 0 due to both gravity and spring potential energy($x=0$).


Now as the block comes down, the spring potential energy is: $U_(spring)=\frac12kx^2$. Final extension is $\frac{mg}k$. So spring potential energy is $\frac{m^2g^2}{2k}$ But the decrease in gravitational potential energy is $mgx$ which equals $\frac{m^2g^2}k$.


This means that potential energy has decreased. Intitially, $U_{net}=0$ but finally $U_{net}=-\frac{m^2g^2}{2k}$.


Where if any, did this energy get compensated(to ensure COE is still true)?




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