135. Let the amount of stretch of the spring be x. For the object to be in equilibrium kxmgxmg k−=¡=0.Thus the gain in elastic potential energy for the spring is ΔUkxkmgkmgke==FHGIKJ=1212222while the loss in the gravitational potential energy of the system is −==FHGIKJ=ΔUmgxmgmgkkgwhich we see (by comparing with the previous expression) is equal to 2ΔUe. The reason why
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.