Question

A conservative force of magnitude $100N$ directed along the line $y=3+x$ is acting on a block of mass $m=1$ kg. Find work done (in J) by given force, when block displaces from point $A(3,1)$ to point B $(1,3)$

Answer 1

As given force is conservative in nature, its work done is dependent only upon initial and final positions.
Since, $y=3+x$
$\therefore tan\theta =\frac{dy}{dx}=1\Rightarrow \theta ={45}^{\circ }$
$\therefore \overrightarrow{F}=100cos{45}^{\circ }\hat{i}+100sin{45}^{\circ }\hat{j}$
$\Rightarrow \overrightarrow{F}=\frac{100}{\sqrt{2}}\hat{i}+\frac{100}{\sqrt{2}}\hat{j}$
The displacement of the particle is
$\overrightarrow{s}=\hat{i}+3\hat{j}-(3\hat{i}+\hat{j})=-2\hat{i}+2\hat{j}$
$\therefore W=\overrightarrow{F}.\overrightarrow{s}=-\frac{200}{\sqrt{2}}+\frac{200}{\sqrt{2}}=0$