ch08-p040 - 40. In this problem, the mechanical energy (the...

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

View Full Document Right Arrow Icon
2 2(16.0 J) 12.6 m/s. 0.200 kg K v m == = (c) At the turning point, the speed of the particle is zero. Let the position of the right turning point be . R x From the figure shown on the right, we find R x to be 16.00 J 0 24.00 J 16.00 J 7.67 m. 7.00 m 8.00 m R RR x xx −− = ¡ = (d) Let the position of the left turning point be . L x From the figure shown, we find L x to be 16.00 J 20.00 J 9.00 J 16.00 J 1.73 m. 1.00 m 3.00 m L LL x = ¡ = 40. In this problem, the mechanical energy (the sum of K and U ) remains constant as the particle moves. (a) Since mechanical energy is conserved,
Background image of page 1
This is the end of the preview. Sign up to 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.

Ask a homework question - tutors are online