The gravitational field in a region is given by `vecE=(10Nkg^-1)(veci+vecj)`. Find the work done by and external agent to slowly shift a particle of mass 2 kg frm the point (0,0) to a point (5m,4m)`

Explanation

As the particle is slowly shifted, its kinetic energy remains zero. The total work done on the particle is ths zero. The work done by the exterN/Al agent should be negative of the work done by th gravitatioN/Al field. The work done by the field is
`int_t^fvecF.vec(dr)` ltbr.gtConsider ure suppose the particle is taken from O to A and then from A to B. The force on the partivle is
`vecF=mvecE=(2kg)(10Nkg^-1)(veci+vecj)=(20N)(veci+vecj)`.

The work done by the field during the displacement OA is
`W_1=int_0^(5m) F_x dx`
`=int_0^(5m)(20N)dx=20Nx5m=100J`
Similarly the work done in displacement AB is
`W_2=int_0^(4m)F_ydy=int_0^(4m)(20N)dy`
`=(20N)(4m)=80J `
Thus,the total work doen by the field as the particle is shifted from O to B is 180 J.
Thework done work is independent of the path so that we can choose any path convenient to us O to B.