Question

A particle is moved along the different paths OAC, OBC & ODC as shown in the figure. Path ODC is a parabola, $y=4x^2.$ If the work done by a force $\overrightarrow{F}=xy\hat{i}+x^{2}y\hat{j}$ on the particle along the path OAC and path OBC are "m" and "n" respectively, then the value of m/n is (Answer upto two digit after decimal point)(x-cordinate of point A is 1 unit. )

Answer 1

From figure, y-cordinate of Point A is zero.Since, x-cordinate of point A is 1 unit , therefore x-cordinate of point c is 1 unit and by equation of parabola, we get y cordinate of point C as 4 unit. Since, C and B have same y-cordinate, we get y-cordinate of point B as 4 and from figure, x-cordinate of point B is zero.

$(W_F{)}_{OAC}=\int (xydx+x^{2}ydy)$
$={\int }_0^A(xydx+x^{2}ydy)$
On OA path;
y = 0, dy = 0 and on AC path x = 1, dx = 0
So $(W_F{)}_{OAC}$
$={\int }_0^A(0.dx+0.dy{)}^{+}{\int }_{y=0}^{y=4}(0+1ydy)=8J=m$
Similarly, For OBC, on OB path x = 0, dx = 0 and on BC path y = 4, dy = 0
$(W_F{)}_{OBC}=0+{\int }_B^C(xydx+x^{2}ydy)$
$={\int }_{x=0}^{x=1}\{x4dx+x^{2}4(0\left)\right\}$
$=2J=n$
Therefore, $\frac{m}{n}=\frac{8}{2}=4$