Match the following

Explanation

Force `vecF=mg(-hatj)=mghatj`
displacement vector `dvecr=dxhati+dxhatj`
(As the displacement is in two dimension ,unit vector `hati and hatj` are used)
(a) Since the motion is only vectrical horizontal displacement component dx is zero. Hence work done by the force along path 1 (of distance h).
`W_("path 1")=underset(A)overset(B)int vecF. d vecr=underset(A)oversetBint (-mg hatj). (dy hatj)=-mg underset(0)overset(h)int dy=-mgh`
Total work done for path 2 is
`W_("path 2")=underset(A)overset(B)int vecF. d vecr=underset(A)oversetC int vecF. d vecr+underset(C)overset(D)int F. d vecr+underset(D)overset(B)int vecF. d vecr`
But `underset(A)overset(C)int vecF. d vecr= underset(A) overset(C)int (- mg hatj). (d x hati)=0`
`underset(C)overset(D)int vecF. d vecr=underset(C)overset(D)int (-mg hatj). (dy hatj)=-mg underset(0)overset(h)int dy=-mgh`
`underset(D)overset(B)int vecF. d vecr=underset(D)overset(B)int (-mg hatj). (-d x hati)=0`
Therefore, the total work done by the force along the path 2 is
`W_("path 2")= undersetA overset(B)int vecF. d vecr=- mgh`
Note that the work done by the conservative force is independent of the path.