Chapter 9283P9.72A picture one second later differs by showing five extra kilograms of sand moving on the belt.(a)∆∆ptx==5000750100375....kgm ssNbgbg(b)The only horizontal force on the sand is belt friction,so from pftpxixf+=∆this isfptx∆∆. N(c)The belt is in equilibrium:Fmaxx∑=:+−=Ffext0andFextN=.(d)WFr°=∆cos.cos.θ0281N 0.750 mJaf(e)12125 000 7501 4122∆mv.kgm sJ(f)Friction between sand and belt converts half of the input work into extra internal energy.*P9.73xmxmmRmmmiiiCM+++=++∑∑12212120AAchafyxRA2FIG. P9.73ANSWERS TO EVEN PROBLEMSP9.2(a) 0; (b) 106kgms⋅; upwardP9.200556mP9.221.78 kN on the truck driver; 8.89 kN in theopposite direction on the car driverP9.4(a) 600.m s to the left; (b) 8.40 J
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