Solve the following :


The gravitational field in a region is given by $E=\left(2 \hat{\imath}+3 j^{\wedge}\right) \mathrm{N} / \mathrm{kg}$. Show that no work is done by the gravitational field when a particle is moved on the line $3 y+2 x=5$

[Hint : If a line $y=m x+c$ makes angle $\theta$ with the $X$-axis, $m=\tan \theta$ ]


Gravitational field in the region is given by

$E=2 \hat{\imath}+3^{\hat{\jmath}}$

Slope of gravitational field, $\mathrm{m}_{1}=\tan \theta_{1}=3 / 2$

The line $3 y+2 x=5$ can be written with the slope, $m_{2}=\tan \theta_{2}=-2 / 3$

$m_{1} \times m_{2}=-1$

Thus, $m_{1}, m_{2}$ lines are perpendicular.

So, no work is done in by gravitational force when a particle is moved on the given line.


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