A force $\vec{F}=(2+3 x) \hat{i}$ acts on a particle in the $x$ direction where F is in newton and $x$ is in meter. The work done by this force during a displacement from $x=0$ to $x=4 \mathrm{~m}$, is __________ J.

<p>To find the work done by a force during a displacement, we can use the formula:</p> <p>$W = \int_{x_1}^{x_2} \vec{F} \cdot d\vec{x}$</p> <p>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:</p> <p>$W = \int_{0}^{4} (2+3x) dx$</p> <p>Now we can integrate the function with respect to $x$:</p> <p>$W = \int_{0}^{4} (2+3x) dx = \int_{0}^{4} 2 dx + \int_{0}^{4} 3x dx$</p> <p>$W = \left[ 2x \right]_0^4 + \left[ \frac{3}{2}x^2 \right]_0^4$</p> <p>Now we can plug in the limits of integration:</p> <p>$$W = (2 \cdot 4 - 2 \cdot 0) + \left(\frac{3}{2} \cdot 4^2 - \frac{3}{2} \cdot 0^2 \right)$$</p> <p>$W = 8 + 24$</p> <p>$W = 32 \mathrm{~J}$</p> <p>So the work done by the force during the displacement from $x = 0$ to $x = 4 \mathrm{~m}$ is 32 Joules.</p>