Constant as eliptical rail `PQ` in the varticle plain with `OP = = 3m ` and `OQ= 4 m` . A block of mass `1` kg is pailed along the rail from `P` to `Q` with a force of `18 N` , which is always parallel to less `PQ` Assuming are frictionless losess , the kinetic energy the block when `0` reches `Q` is `( nxx 10)` pales . THe velie of a (Take acceleration due to gravity ) `= 10ms^(-2))`

Explanation

As is clear from figure,
`PQ=sqrt(OP^(2)+OQ)=sqrt(3^(2) + 4^(2))=5 m`
Work done is taking the block from `P` to `Q` along elliptically shaped rail,.
`W=FxxPQ=18 xx 5 =90J`
Potential energy of block at `Q=+ mgh`
`=+1 xx 10 xx 4 =+40J`
`:. K.E. ` at `Q=W-P.E.=90-40=50J=10n`
`:. n=5`