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A solid circular disc of mass 50 kg rolls along a horizontal floor so that its center of mass has a speed of 0.4 m/s. The absolute value of work done on the disc to stop it is J.
6
Work–energy theorem: W= K=0-K
Rolling disc: K= 12mv2+ 12I 2 with I= 12mR2, = vR
K= 12mv2 (1+ 12 )= 34mv2
= 34 50 (0.4)2=6 J
|W|=6 J
Answer: 6
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