QJEEM25Ph23b7: JEE Main 2025 Physics - Class 11 JEEM25Ph23b7 Solution: A force acts on a particle in a plane . The work done by this…

To calculate the work done by the force $\mathrm{f} = \mathrm{x}^2 \mathrm{y} \hat{\mathrm{i}} + \mathrm{y}^2 \hat{\mathrm{j}}$ during the displacement from $(0,0)$ to $(4 \text{ m}, 2 \text{ m})$ , we proceed as follows:

Given the constraint that the particle moves in the plane $\mathrm{x} + \mathrm{y} = 10$ , we can express $\mathrm{y}$ in terms of $\mathrm{x}$ :

$y = 10 - x$

The work done, $W$ , is the integral of the force along the path of the displacement:

$W = \int_{0}^{4} x^2 (10-x) \, dx + \int_{0}^{2} y^2 \, dy$

Calculate the integral of the $\hat{\mathrm{i}}$ component:

$\begin{aligned} & \int_{0}^{4} x^2 (10 - x) \, dx = \int_{0}^{4} (10x^2 - x^3) \, dx \\ & = \left[ \frac{10x^3}{3} - \frac{x^4}{4} \right]_{0}^{4} \\ & = \left( \frac{10(4)^3}{3} - \frac{(4)^4}{4} \right) - \left( \frac{10(0)^3}{3} - \frac{(0)^4}{4} \right) \\ & = \frac{640}{3} - \frac{256}{4} \\ \end{aligned}$

Calculate the integral of the $\hat{\mathrm{j}}$ component:

$\begin{aligned} & \int_{0}^{2} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{2} \\ & = \frac{(2)^3}{3} - \frac{(0)^3}{3}\\ & = \frac{8}{3} \\ \end{aligned}$

Combine the results:

$W = \left( \frac{640}{3} - 64 \right) + \frac{8}{3}$

Simplify:

$\begin{aligned} W & = \frac{640}{3} - 64 + \frac{8}{3} \\ & = \frac{640 + 8}{3} - 64 \\ & = \frac{648}{3} - 64 \\ & = 216 - 64 \\ & = 152 \text{ J} \end{aligned}$

Thus, the work done by the force is 152 Joules.