Then wnc f22 em 11u 11k ikx2 i690 nm014 m2

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: quot;;kx2/m = = = = = = Problem II ";(690 N/m) (0.14 m)2j(0.05 kg) 16.4 mis, so the time it takes to travel 21 m (horizontally) is approximately t = d/v= 21 m/(16.4 m/s) = 1.28 s. = 57. A surface is frictionless except for a region between x = 1 m, and x ::::: m, where thecoefflcient of 2 friction is given by I-' = ax2 + bx + c, v.lith 2 b = 6 m-I, and c = -4. A block is a:::: -2 m- , sliding in the -x direction when it encbunters this region. What is the minimum speed it li!musthave to get all the way across the region? II Problem 59. A skier starts from rest at the topofthe left-hand peak in Fig. 8-39. What is the maximum coefficient of kinetic friction on the slopes that would allow the skier to coast to the second peak? (Your answer, of course, neglects air resistance.) Solution II Solution Assume that the maximal frictional force of the slopes and gravity are the only significant forces acting on the skier, who starts from rest on the higher peak (point A) and finishes at rest just reaching the lower peak (point B). Then the work-energy theorem requires that Wr = 11U :'::: mg(YB ~ YA)' The force of friction on each slope is !k = - JJ.r:ax mg cos 8, and the displ...
View Full Document

Ask a homework question - tutors are online