The rotational kinetic energy of a solid sphere of  mass 3 kg and radius 0.2 m rolling down an inclined plane of height 7 m is :

mgh=KET  +  KER=12 mv2  +  12  Iω2=12  mv2 + 12  25  mR2  v2R2=12  mv2 + 15mv2=  710  mv2mv2 =107 ×  3  ×  10  ×  7=300kER= 12  Iω2=15  mv2=60J