QJEEM23Ph2881: JEE Main 2023 Physics - Class 11 JEEM23Ph2881 Solution: A block of mass is moving along -axis under the action of force …

To calculate the work done by the force $F = 5x$ in moving the block from $x = 2m$ to $x = 4m$ , we can use the formula for work done by a variable force:

$W = \int_{x_1}^{x_2} F(x) dx$

In this case, $F(x) = 5x$ , $x_1 = 2m$ , and $x_2 = 4m$ . Now, we can substitute these values into the formula and evaluate the integral:

$W = \int_{2}^{4} 5x dx$

To evaluate the integral, we find the antiderivative of $5x$ :

$\int 5x dx = \frac{5}{2}x^2 + C$

Now, we can find the work done by evaluating the antiderivative at the limits of integration:

$W = \left[\frac{5}{2}x^2\right]_{2}^{4} = \frac{5}{2}(4^2) - \frac{5}{2}(2^2)$

$W = \frac{5}{2}(16) - \frac{5}{2}(4) = 40 - 10 = 30 \mathrm{J}$

The work done by the force in moving the block from $x = 2m$ to $x = 4m$ is 30 J.