Kinetic energy of a rolling hoop

[Jrlinton]

Homework Statement

A 120 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.240 m/s. How much work must be done on the hoop to stop it?

Homework Equations

I of hoop=MR^2

The Attempt at a Solution

KE=0.5*m*v^2+0.5*(mR^2)(v/R)^2
=0.5*120kg*.24^2m/s+0.5*(120kg*R^2)(.24m/s^2/R^2)
=6.912 J

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[gneill]

Your calculation looks good. You might want to trim the significant digits in your result to match the given data. If this is a "show your work" type question rather than a web based "enter the solution" assignment, you might want to include a note as to why the total KE is the same as the work required.