Question

The gravitational field in a region due to a certain mass distribution is given by $\overrightarrow{E}=(4\hat{i}-3\hat{j})\text{N/kg}.$ The work done by the field in moving a particle of mass $2\text{kg}$ from $(2\text{m},1\text{m})$ to $(\frac{2}{3}\text{m},2\text{m})$ along the line $3x+4y=10$ is

• A. $-\frac{25}{3}\text{J}$
• B. $-\frac{50}{3}\text{J}$
• C. $\frac{25}{3}\text{J}$
• D. Zero

Answer 1

The correct option is B $-\frac{50}{3}\text{J}$

Given that,
Gravitational field, $\overrightarrow{E}=(4\hat{i}-3\hat{j})\text{N/kg}$

Initial position of partical, $P_i=(2\text{m},1\text{m})$

Final position of particle, $P_f=(\frac{2}{3}\text{m},2\text{m})$

Equation of line is $3x+4y=10$

The relation between gravitational field and potetial is
Potential Difference, $dV=-\int \overrightarrow{E}.\overrightarrow{dr}$

Since $\overrightarrow{E}$ is constant,

$\Delta V=-\overrightarrow{E}.\int \overrightarrow{dr}=-\overrightarrow{E}.\overrightarrow{r}$
$\overrightarrow{E}=(4\hat{i}-3\hat{j})\text{Nkg}$
$\overrightarrow{r}=\left(\right(\frac{2}{3}-2)\hat{i}+(2-1\left)\hat{j}\right)\text{m}$
$\overrightarrow{r}=(-\frac{4}{3}\hat{i}+\hat{j})\text{m}$
$\Rightarrow -\overrightarrow{E}.\overrightarrow{r}=(-4\hat{i}+3\hat{j}).(-\frac{4}{3}\hat{i}+\hat{j})$
$\Rightarrow \Delta V=-\overrightarrow{E}.\overrightarrow{r}=\frac{16}{3}+3$
$\Rightarrow \Delta V=-\overrightarrow{E}.\overrightarrow{r}=\frac{25}{3}$

The work done by gravitational force does not depend on path, it depends on the initial and final position only.

Work done = Change in potential energy
$m\Delta V=2\times \frac{25}{3}=\frac{50}{3}\text{J}$

Key Concept: When field vector is constant, $V=-\overrightarrow{E}.\overrightarrow{r}$