Question

A rod of mass `m` and length `l` is lying on a horizontal table. Work done in making it stand on one end will be
A. `mgl`
B. `mgl//2`
C. `(mgl)/(4)`
D. `2 mgl`

Answer 1

Here, mass of rod `=m, ` length`-l`
Work done `=` final `P.E. -` length `P.E.`
`P_(i)=0[` as it is lying on horizontal table,
`:. H=0` and `P.E.=mgh=0]`
`P_(f)=mg((l)/(2)f)`
Work done `=P_(f)-P_(i)=mg ((l)/(2))-0=mg((l)/(2))`