College Physics (4th Edition)

The work done on the stone is $151~J$
We can find the work done on the stone: $Work = \tau ~\Delta \theta$ $Work = r~F ~\Delta \theta$ $Work = (0.10~m)(20.0~N)(12~rev)(2\pi~rad/rev)$ $Work = 151~J$ The work done on the stone is $151~J$.