QJEEM24Ph8245: JEE Main 2024 Physics - Class 11 JEEM24Ph8245 Solution: A force displaces a body from to . Work done by this force is…

To find the work done by the force when the body is displaced from $x = 2 \, \mathrm{m}$ to $x = 4 \, \mathrm{m}$ , we use the formula for work done by a variable force in one dimension, which is the integral of the force with respect to displacement:

$W = \int_{x_1}^{x_2} F \, dx$

Given the force $F(x) = (3x^2 + 2x - 5) \, \mathrm{N}$ and the limits of integration from $x = 2 \, \mathrm{m}$ to $x = 4 \, \mathrm{m}$ , we can substitute these values into the equation:

$W = \int_{2}^{4} (3x^2 + 2x - 5) \, dx$

Calculating the integral, we get:

$W = \left[\frac{3x^3}{3} + \frac{2x^2}{2} - 5x\right]_2^4$

This simplifies to:

$W = \left[x^3 + x^2 - 5x\right]_2^4$

Substituting the upper limit ( $x = 4$ ) and then the lower limit ( $x = 2$ ) into the antiderivative, and subtracting the latter from the former, we get:

$W = \left[(4)^3 + (4)^2 - 5(4)\right] - \left[(2)^3 + (2)^2 - 5(2)\right]$

$W = (64 + 16 - 20) - (8 + 4 - 10)$

$W = 60 - 2$

$W = 58 \, \mathrm{J}$

Therefore, the work done by the force as the body displaces from $x = 2 \, \mathrm{m}$ to $x = 4 \, \mathrm{m}$ is $58 \, \mathrm{J}$ .