A 190 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.160 m/s. How much work must be done on the hoop to stop it?
Physics - Elena, Wednesday, April 18, 2012 at 6:32pm
W =KE = m•v^2/2 + I•ω^2/2 =
= m•v^2/2 +(mR^2) •(v^2/R)/2= mv^2 =
=190•0.16^2 =4.86.4 J