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 :

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Detailed Solution

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

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