QJEEM23Ph19cb: JEE Main 2023 Physics - Class 11 JEEM23Ph19cb Solution: A force acts on a particle in the direction where F is in…

To find the work done by a force during a displacement, we can use the formula:

$W = \int_{x_1}^{x_2} \vec{F} \cdot d\vec{x}$

Here, the force is given by $\vec{F} = (2+3x) \hat{i}$ , and we need to find the work done during a displacement from $x = 0$ to $x = 4 \mathrm{~m}$ . Since the force is only in the $x$ direction, we can write the integral as:

$W = \int_{0}^{4} (2+3x) dx$

Now we can integrate the function with respect to $x$ :

$W = \int_{0}^{4} (2+3x) dx = \int_{0}^{4} 2 dx + \int_{0}^{4} 3x dx$

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

Now we can plug in the limits of integration:

$W = (2 \cdot 4 - 2 \cdot 0) + \left(\frac{3}{2} \cdot 4^2 - \frac{3}{2} \cdot 0^2 \right)$

$W = 8 + 24$

$W = 32 \mathrm{~J}$

So the work done by the force during the displacement from $x = 0$ to $x = 4 \mathrm{~m}$ is 32 Joules.