A body of mass 500 gram is rotating in a vertical of radius 1 m . What is the difference in its kinetic energies at the top and the bottom of the circle ?
A
`4.9 J`
B
`19.8 J`
C
`2.8 J`
D
`-9.8 J`
Text Solution
Verified by Experts
The correct Answer is:
B
Diff .In . K.E at the bottom and at the top ` =(1)/(2) m (v_(1)^(2)-v_(2)^(2)) "But " v_(1) sqrt(5 gr ) and v_(2) = sqrt( gr)` `=(1)/(2) m ( 5 gr 0 gr) = 2 mgr = 2xx(1)/(2) xx9.8 xx1` `=9.8J`
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