sample2_extra_exam_soln - Solutions to sample exam 2 TAM...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Solutions to sample exam 2 TAM 212, Fall, 2002 Solution for sample problem #1 1 : The change in kinetic energy of the plunger = work done by external forces: Work E K = . . () () 2 2 2 / 1 2 / 1 . . initial plunger final plunger v m v m E K = 0 = initial v , so () ( ) ( ) 2 2 / 32 / 5 2 / 1 . . v s ft lb E K = , where v is its speed just before it hits the button. The external forces include that due to gravity and that due to the spring. () ( ) lb in lb in h mg Work down gravity = = = 75 . 8 75 . 1 5 () ( ) 2 2 2 / 1 final initial sping k Work δ δ = To find the δ ’s we need to solve for it in the equilibrium position: () in lb in lb F F ma F gravity spring y y 5 . 0 5 / 10 0 = δ = δ = + = so () ( ) ( )( ) ( ) lb in in in in lb Work sping = == 1875 . 2 75 . 0 1 / 10 2 / 1 2 2 () () () () () ( ) 2 2 2 2 2 / 667 . 11 12 / 1 / 140 1875 . 2 75 . 8 / / 078125 . s ft in ft x s ft in v lb in v s ft lb = = + =
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/09/2008 for the course TAM 212 taught by Professor Keane during the Spring '08 term at University of Illinois at Urbana–Champaign.

Page1 / 2

sample2_extra_exam_soln - Solutions to sample exam 2 TAM...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online