Post a New Question


posted by .

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 -

    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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question