07_InstSolManual_PDF_Part13

07_InstSolManual_PDF_Part13 - 7.60. Set Up: Applying with...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 7.60. Set Up: Applying with the change in mechanical energy is simply the differ- ence between the initial and final elastic potential energy of the stretched spring: Also define the initial position as the point of release and the final position as the lowest position after five complete cycles. Figure 7.60 Solve: (a) (b) The percentage lost is Reflect: Note that the spring, in its equilibrium position, is stretched by the weight of the hanging mass. Conse- quently, an alternative solution can be developed that explicitly incorporates gravitational potential energy. Let d be the distance the spring is stretched at equilibrium due to the weight of the object. Define a new reference point; let where the mass is 0.130 m below the equilibrium point. At this point, the spring is stretched The total potential energy is therefore: Then when the mass is 0.100 m below the equilibrium position, and the spring is stretched The total energy is then: The change in mechanical energy is thus the same as found in the simplified calculation of...
View Full Document

This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

Ask a homework question - tutors are online