A chain of mass m and length l is overhinges a table such that its two third part is lying on table.Find the K.E of chain as it completely slips offf the table?
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W.D in slipping of chain = ∆KE
here initially KE = 0 so we have to calculate KE final
now W.D. = mgl/ 2n× n where n is the hanging part
=mgl/2×{(2/3)×(2/3)} = 9mgl/8 JOULE
here initially KE = 0 so we have to calculate KE final
now W.D. = mgl/ 2n× n where n is the hanging part
=mgl/2×{(2/3)×(2/3)} = 9mgl/8 JOULE
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