A bead slides along a frictionless wire lying on a horizontal plane. It makes an anlge of `45^(@)` with x-axis as show in figure. In addition to any normal forces exerted by the wire, the bead is subject to an external force that depends on position according to formula.
`vec(F)=F_(0)((x)/(x_(0)))^(2)hat(i)+F_(0)((y)/(y_(0)))^(2)hat(j)`
If the bead has speed `v_(0)` at the origin O, find its speed (in m/s) as it leaves the end P of wire Given data : `F_(0)=1N, x_(0)=15m, y_(0)=15m, " mass of bead"=2kg` Initially kinetic energy of bead = 26 J

Answer

6

Explanation

`WD_(F)=intvec(F).dvec(r )`
`=(F_(0))/(x_(0)^(2))underset(0)overset(x_(0))intx^(2)dx+(F_(0))/(y_(0)^(2))underset(0)overset(y_(0))inty^(2)dy`
`=(F_(0))/(3)(x_(0)+y_(0))=10J`
`WD=k_(l)-k_(L)`
`10 " "=k_(l)-26 " implies "k_(l)=36`
`" implies "(1)/(2)xx2xxv^(2)=36`
`" "therefore" "v=6m//s`