91Ó°ÊÓ

We know that work done by a varying force\(\overrightarrow F \)on a particle moving along a path from point A to B is given by

\(W = \int_A^B {\overrightarrow {F \cdot } \overrightarrow {dl} } \) ...(1)

Where\(\overrightarrow {dl} \)is the infinitesimal displacement.

we have given that varying force is\(\overrightarrow F = - \alpha x{y^2}\widehat {j\;}\;{\rm{where }}\alpha = 2.50\,{\rm{N/}}{{\rm{m}}^{\rm{3}}}\)

And we have to consider displacement of the tool from the origin to the point \(\left( {x = 3\,{\rm{m}},\;y = 3\,{\rm{m}}} \right)\).

Here, we have the initial point as\(\left( {{x_1},{y_1}} \right) = \left( {0,0} \right)\) and the final point is \(\left( {{x_2},{y_2}} \right) = \left( {3,3} \right)\).

Now, displacement is along a straight line \(y = x\),

Then we have \(\overrightarrow {dl} = dx\widehat i + dy\widehat j\)

Also we have, \(A = \left( {0,0} \right)\) and \(B = \left( {3,3} \right)\)

So, by equation (1)

\(W = \int_{\left( {0,0} \right)}^{\left( {3,3} \right)} { - \alpha x{y^2}\widehat {j\;}} \cdot \left( {dx\widehat i + dy\widehat j} \right)\)

\(W = \int_0^3 { - \alpha x{y^2}dy} \)

Also we have, \(y = x\)

\(\begin{aligned}{}W &= - \alpha \int_0^3 {{y^3}dy} \\W &= - \alpha \left( {\frac{{{y^4}}}{4}} \right)_0^3\end{aligned}\)

Substitute all the value in the above equation.

\(\begin{aligned}{}W &= - 2.5\,{\rm{N/}}{{\rm{m}}^{\rm{3}}}\left( {\frac{{{{\left( {3\,{\rm{m}}} \right)}^4}}}{4} - 0} \right)\\W &= - 50.625\,{\rm{J}}\end{aligned}\)

Hence the work done on the tool by \(\overrightarrow F \) if this displacement is along the straight line \(y = x\) that connects these two points is \( - 50.625\,{\rm{J}}\).

Here first we will find the work done for moved out along the x-axis to the point\(\left( {3,0} \right)\).

We have the initial point as\(\left( {{x_1},{y_1}} \right) = \left( {0,0} \right)\)and the final point is \(\left( {{x_2},{y_2}} \right) = \left( {3,0} \right)\)

Here it moved along the x-axis.

So, infinitesimal displacement is given by

.\(\overrightarrow {dl} = dx\widehat i\)

Now, from equation (1)

\({W_1} = \int_{\left( {0,0} \right)}^{\left( {3,0} \right)} { - \alpha x{y^2}\widehat j} \cdot \left( {dx\widehat i} \right)\)

\({W_1} = 0\)

Now, we will find work done for moved parallel to y-axis to the point \(\left( {3,3} \right)\)

Here now we the initial point as\(\left( {{x_1},{y_1}} \right) = \left( {3,0} \right)\)and the final point is \(\left( {{x_2},{y_2}} \right) = \left( {3,3} \right)\)

Here it moves parallel to y-axis.

So, infinitesimal displacement is given by

.\(\overrightarrow {dl} = dy\widehat j\)

Now, from equation (1)

\({W_1} = \int_{\left( {3,0} \right)}^{\left( {3,3} \right)} { - \alpha x{y^2}\widehat {j\;}} \cdot \left( {dy\widehat j} \right)\)

\(W = \int_0^3 { - \alpha x{y^2}dy} \)

Here the particle moving only parallel to y-axis. So, \(x = 3\)

\(W = \int_0^3 { - \alpha 3{y^2}dy} \)

\(\begin{aligned}{}{W_2} &= - 3\alpha \int_0^3 {{y^2}dy} \\{W_2} &= - 3\alpha \left( {\frac{{{y^3}}}{3}} \right)_0^3\end{aligned}\)

Substitute all the value in the above equation.

\(\begin{aligned}{}{W_2} &= - 3 \times 2.5\,{\rm{N/}}{{\rm{m}}^{\rm{3}}}\left( {\frac{{{{\left( {3\,{\rm{m}}} \right)}^3}}}{3} - 0} \right)\\{W_2} &= - 67.5\,{\rm{J}}\end{aligned}\)

Now, total work is,

\(\begin{aligned}{}W &= {W_1} + {W_2}\\ = 0 - 67.5\,{\rm{J}}\\W &= - 67.5\,{\rm{J}}\end{aligned}\)

Hence the work done on the tool by \(\overrightarrow F \) if the tool is first moved out along the x-axis to the point \(\left( {x = 3.00{\rm{ m}},y = 0} \right)\) and then moved parallel to the y-axis to the point\(\left( {x = 3.00\,{\rm{m}},y = 3.00\,{\rm{m}}} \right)\)is \( - 67.5\,{\rm{J}}\).

Here we can see that in (a) and (b), initial point and final point are same but they have different values of work done.

So, we can say that \(\overrightarrow F = - \alpha x{y^2}\widehat {j\;}\) is nonconservative.