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NVAMediumJEE 2025Work Done by Force

JEE Physics 2025 Question with Solution

A force f⃗=x2i^+yj^+y2k^\vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} acts on a particle in a plane x+y=10x + y = 10. The work done by this force during a displacement from (0,0)(0,0) to (4m,2m)(4m, 2m) is _____ Joules (round off to the nearest integer).

Answer

Correct answer:152

Step-by-step solution

Standard Method

Given: f⃗=x2i^+yj^+y2k^\vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} and the motion is constrained by x+y=10x + y = 10.

Find: The work done during the displacement from (0,0)(0,0) to (4,2)(4,2).

From the extracted solution, the work is evaluated as:

W=∫04x2(10−x) dx+∫02y2 dyW=\int_0^4 x^2(10-x)\,dx + \int_0^2 y^2\,dy

Now evaluate the first integral:

∫04(10x2−x3) dx=[10x33−x44]04=6403−64=4483\int_0^4 (10x^2 - x^3)\,dx =\left[\frac{10x^3}{3}-\frac{x^4}{4}\right]_0^4 =\frac{640}{3}-64=\frac{448}{3}

Evaluate the second integral:

∫02y2 dy=[y33]02=83\int_0^2 y^2\,dy =\left[\frac{y^3}{3}\right]_0^2 =\frac{8}{3}

Combine the two results:

W=4483+83=4563=152W=\frac{448}{3}+\frac{8}{3}=\frac{456}{3}=152

Therefore, the work done is 152 J152 \, \text{J}. Hence, the required numerical answer is 152.

Stepwise Evaluation

Given: The force field is f⃗=x2i^+yj^+y2k^\vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k}.

Find: The total work done.

The extracted solution splits the line integral into two parts:

W=∫04x2(10−x) dx+∫02y2 dyW = \int_0^4 x^2(10 - x)\,dx + \int_0^2 y^2\,dy

For the first part,

∫04x2(10−x) dx=∫04(10x2−x3) dx\int_0^4 x^2(10-x)\,dx = \int_0^4 (10x^2 - x^3)\,dx =[10x33−x44]04= \left[ \frac{10x^3}{3} - \frac{x^4}{4} \right]_0^4 =10(43)3−444= \frac{10(4^3)}{3} - \frac{4^4}{4} =6403−64=4483= \frac{640}{3} - 64 = \frac{448}{3}

For the second part,

∫02y2 dy=[y33]02=83\int_0^2 y^2\,dy = \left[ \frac{y^3}{3} \right]_0^2 = \frac{8}{3}

Adding them,

W=4483+83=4563=152W = \frac{448}{3} + \frac{8}{3} = \frac{456}{3} = 152

Therefore, the correct numerical value is 152.

Common mistakes

  • Using the answer key 12 instead of the value concluded by the solution. The solution is the primary source here and it clearly gives 152 J152 \, \text{J}. Always derive the answer from the worked solution when available.

  • Mis-evaluating the polynomial integral ∫04(10x2−x3) dx\int_0^4 (10x^2 - x^3)\,dx. This is not equal to a small number like 1212; the cubic and quartic contributions are substantial over the interval. Expand first, integrate term by term, and then substitute the limits carefully.

  • Ignoring that the extracted solution splits the work into two separate integrals in xx and yy. If one term is omitted, the final work will be incorrect. Follow the same decomposition shown in the provided working and add both contributions.

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