Question

The gravitational field in a region is given by $\overrightarrow{E}=(3\hat{i}-4\hat{j})$N $k{g}^{-1}$. Find out the work done(in joule) in displacing a particle by $1$m along the line $4y=3x+9$.

Answer 1

We have $\overrightarrow{E}=(3\hat{i}-4\hat{j})N/kg$

Force acting as unit mass $\overrightarrow{F}=m\overrightarrow{E}$

$=(3\hat{i}-4\hat{j})N$

The motion of unit mass along the line $4y=3x+9$

$y=\frac{3x+9}{4}$

At $x_2=1$ $y_2=\frac{12}{4}=3$ let the point be $A$.

At $x_1=0$ $y_1=\frac{9}{4}$

So vector $\overrightarrow{d}$ from $A$ to $B$

$\overrightarrow{d}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}$

$=(1-0)\hat{i}+\frac{(3-9)}{4}\hat{j}$

$=\hat{i}+\left(\frac{12-9}{4}\right)\hat{j}$

$=\hat{i}+\frac{3}{4}\hat{j}$

Now work done is moving unit mass a long given line is

$W=\overrightarrow{F}.\overrightarrow{d}=(3\hat{i}-4\hat{j}).(\hat{i}+\frac{3}{4}\hat{j})$

$=3\hat{i}\times \hat{i}-4\hat{j}\times \frac{3}{4}\hat{j}$

$=3-3=0$

Thus $F\perp d$ hence work done is moving along given line is zero.