The gravitational field in a region is given by $\vec{E}=\left(10 \mathrm{~N} \mathrm{~kg}^{-1}\righ

Question

The gravitational field in a region is given by $\vec{E}=\left(10 \mathrm{~N} \mathrm{~kg}^{-1}\right)(\vec{i}+\vec{j})$. The work done by an external agent to slowly shift a particle of mass 2 kg from the point $(0,0)$ to a point $(5 \mathrm{~m}, 4 \mathrm{~m})$ is

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As the particle is slowly shifted, its kinetic energy remains zero. The total work done on the particle is thus zero. The work done by the external agent should be negative of the work done by the gravitational field. The work done by the field is\ $\int_i^f \vec{F} \cdot d \vec{r}$ \The work is independent of the path so that we can choose any path convenient to us from O to B. Suppose the particle is taken from O to A and then from A to B. The force on the particle is $\vec{F}=m \vec{E}=(2 \mathrm{~kg})\left(10 \mathrm{~N} \mathrm{~kg}^{-1}ight)(\vec{i}+\vec{j})=(20 \mathrm{~N})(\vec{i}+\vec{j})$
The work done by the field during the displacement OA is
$\begin{aligned} W_1 & =\int_0^{5 m} F_x d x \ & =\int_0^{5 m}(20 \mathrm{~N}) d x=20 \mathrm{~N} 	imes 5 \mathrm{~m}=100 \mathrm{~J}\end{aligned}$
Similarly, the work done in displacement AB is
$\begin{aligned} W_2 & =\int_0^{4 m} F_y d y=\int_0^{4 m}(20 \mathrm{~N}) d y \ & =(20 \mathrm{~N})(4 \mathrm{~m})=80 \mathrm{~J}\end{aligned}$
Thus, the total work done by the field, as the particle is shifted from $O$ to $B$, is 180 J. The work done by the external agent is -180 J .