A person brings a mass of 1 kg from infinity to a point A. Initially the mass was at rest but it moves with a speed of 2 m/s as it reaches A. The work done by the person on the mass is – 3 J. The potential of A is

The correct option is – 5 J/kg.

To find the gravitational potential at point A, we can apply the work-energy theorem.
The work-energy theorem states that the total work done on an object by all forces is equal to the change in its kinetic energy:
$W_{\text{total}}=\Delta K$

Here, the forces acting on the mass are the external force exerted by the person ($W_{\text{person}}$) and the gravitational force ($W_{\text{gravity}}$). Therefore:
$W_{\text{person}}+W_{\text{gravity}}=K_{\text{final}}-K_{\text{initial}}$

Let's list the given values:
Mass, $m=1\text{kg}$
Initial speed at infinity, $v_{\text{initial}}=0\text{m/s}$ (since it was initially at rest)
Final speed at point A, $v_{\text{final}}=2\text{m/s}$
Work done by the person, $W_{\text{person}}=-3\text{J}$

First, we calculate the change in kinetic energy ($\Delta K$):
$K_{\text{initial}}=0$
$K_{\text{final}}=\frac{1}{2}m{v}_{\text{final}}^2=\frac{1}{2}·1·2^2=2\text{J}$
$\Delta K=2\text{J}-0\text{J}=2\text{J}$

Now, substitute the values into the work-energy relation:
$-3+W_{\text{gravity}}=2$
$W_{\text{gravity}}=2+3=5\text{J}$

The work done by the gravitational field is related to the change in gravitational potential energy ($\Delta U$) by:
$W_{\text{gravity}}=-\Delta U=-(U_{\text{A}}-U_{\infty })$

Since the potential energy at infinity ($U_{\infty }$) is taken as zero:
$W_{\text{gravity}}=-U_{\text{A}}$
$5\text{J}=-U_{\text{A}}\Rightarrow U_{\text{A}}=-5\text{J}$

Gravitational potential ($V_{\text{A}}$) at point A is defined as the potential energy per unit mass:
$V_{\text{A}}=\frac{U_{\text{A}}}{m}=\frac{-5\text{J}}{1\text{kg}}=-5\text{J/kg}$