Questions
A particle of mass 'm' is moving in a horizontal circle of radius 'r' under a centripetal force equal to
detailed solution
Correct option is B
TE=KE+PE KE=12mv2(F=mv2r=kr2) 12×kr=k2r PE=∫∞rFdr=−kr TE=k2r−kr=−k2rTalk to our academic expert!
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