Work and Rotational Kinetic Energy

[G-reg]

Homework Statement

A 38.0 kg wheel, essentially a thin hoop with radius 1.13 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.
(a) How much work must be done to stop it?

Homework Equations

change in KE = (1/2)Iwi^2 - (1/2)Iwf^2
I = (1/2)mr^2

The Attempt at a Solution

(1/2)[(1/2)(38)(1.13)^2](28pi/3) = 355.686 kg(m)^2/s

I don't see where the time comes into play though..can anyone help?

 

[rl.bhat]

(1/2)[(1/2)(38)(1.13)^2](28pi/3)
It should be
(1/2)[(1/2)(38)(1.13)^2](28pi/3)^2

To find the work done, time is not necessary.

 

[G-reg]

Oh ok, I get it now!
Thank you so much!