A force f=x^2 y \hati+y^2 \hatj acts on a particle in a plane x+y=10 . The work done by this force during a displacement from (0,0) to (4 ~m, 2 ~m) is \underline\hspace2cm Joule (round off to the nearest integer)

Work Power and Energy PYQ — JEE Main 2025 · 23 Jan · Shift 1 · Q72 with Solution

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Correct answer: 35

Given force field

The force is

\vec F = x^2 y\,\hat i + y^2\,\hat j

So,

F_x = x^2 y, \qquad F_y = y^2

We need the work done in moving the particle from $(0,0)$ to $(4,2)$ .

Equation of path

The particle moves in the plane curve

x+y=10

But the given initial point $(0,0)$ and final point $(4,2)$ do not satisfy $x+y=10$ .

So the path condition is inconsistent with the endpoints. In such questions, the intended path is almost certainly the straight line joining $(0,0)$ to $(4,2)$ , whose equation is

y=\frac{x}{2}

We proceed with this physically consistent path.

Expression for work

Work done is

W=\int \vec F\cdot d\vec r = \int (F_x\,dx + F_y\,dy)

Along the path

y=\frac{x}{2} \implies dy = \frac{1}{2}dx

Also, limits are:

x:0\to 4

Substitute into the integrand

First,

F_x = x^2 y = x^2\left(\frac{x}{2}\right)=\frac{x^3}{2}

and

F_y = y^2 = \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}

Therefore,

W=\int_0^4 \left(F_x\,dx + F_y\,dy\right)

W=\int_0^4 \left(\frac{x^3}{2}\,dx + \frac{x^2}{4}\cdot \frac{1}{2}dx\right)

W=\int_0^4 \left(\frac{x^3}{2}+\frac{x^2}{8}\right)dx

Integrate

W=\int_0^4 \frac{x^3}{2}\,dx + \int_0^4 \frac{x^2}{8}\,dx

W=\frac{1}{2}\left[\frac{x^4}{4}\right]_0^4 + \frac{1}{8}\left[\frac{x^3}{3}\right]_0^4

W=\frac{1}{8}(4^4) + \frac{1}{24}(4^3)

W=\frac{256}{8} + \frac{64}{24}

W=32 + \frac{8}{3}

W=\frac{104}{3} \approx 34.67\text{ J}

Rounded to nearest integer:

\boxed{35}

Comparison with stored answer

Stored correct answer is $152$ , but the consistent evaluation gives

\boxed{35}

Hence, I do not agree with the stored answer. The likely reason is that the path equation in the question is misprinted or the stored answer corresponds to some different path/data.

More from Work Power and Energy

View written solutionFree

Correct answer: 35

Given force field

The force is

\vec F = x^2 y\,\hat i + y^2\,\hat j

So,

F_x = x^2 y, \qquad F_y = y^2

We need the work done in moving the particle from $(0,0)$ to $(4,2)$ .

Equation of path

The particle moves in the plane curve

x+y=10

But the given initial point $(0,0)$ and final point $(4,2)$ do not satisfy $x+y=10$ .

So the path condition is inconsistent with the endpoints. In such questions, the intended path is almost certainly the straight line joining $(0,0)$ to $(4,2)$ , whose equation is

y=\frac{x}{2}

We proceed with this physically consistent path.

Expression for work

Work done is

W=\int \vec F\cdot d\vec r = \int (F_x\,dx + F_y\,dy)

Along the path

y=\frac{x}{2} \implies dy = \frac{1}{2}dx

Also, limits are:

x:0\to 4

Substitute into the integrand

First,

F_x = x^2 y = x^2\left(\frac{x}{2}\right)=\frac{x^3}{2}

and

F_y = y^2 = \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}

Therefore,

W=\int_0^4 \left(F_x\,dx + F_y\,dy\right)

W=\int_0^4 \left(\frac{x^3}{2}\,dx + \frac{x^2}{4}\cdot \frac{1}{2}dx\right)

W=\int_0^4 \left(\frac{x^3}{2}+\frac{x^2}{8}\right)dx

Integrate

W=\int_0^4 \frac{x^3}{2}\,dx + \int_0^4 \frac{x^2}{8}\,dx

W=\frac{1}{2}\left[\frac{x^4}{4}\right]_0^4 + \frac{1}{8}\left[\frac{x^3}{3}\right]_0^4

W=\frac{1}{8}(4^4) + \frac{1}{24}(4^3)

W=\frac{256}{8} + \frac{64}{24}

W=32 + \frac{8}{3}

W=\frac{104}{3} \approx 34.67\text{ J}

Rounded to nearest integer:

\boxed{35}

Comparison with stored answer

Stored correct answer is $152$ , but the consistent evaluation gives

\boxed{35}

Hence, I do not agree with the stored answer. The likely reason is that the path equation in the question is misprinted or the stored answer corresponds to some different path/data.

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