Potential energy function corresponding to a conservative force is given as U(x, y, z) = 3x²/2 + 5y + 6z, then the force at x = 6 is pN. The value of p upto its nearest integral value is

The correct answer is 20.

To find the force associated with a given conservative potential energy function, we use the relationship between conservative force and potential energy:
$\overrightarrow{F}=-\nabla U$
where $\nabla U$ is the gradient of the potential energy function $U(x,y,z)$.

The potential energy function is given by:
$U(x,y,z)=\frac{3x^2}{2}+5y+6z$

First, we calculate the partial derivatives of $U$ with respect to $x$, $y$, and $z$ to find the components of the force vector:
$F_x=-\frac{\partial U}{\partial x}=-\frac{\partial }{\partial x}(\frac{3x^2}{2}+5y+6z)=-3x$
$F_y=-\frac{\partial U}{\partial y}=-\frac{\partial }{\partial y}(\frac{3x^2}{2}+5y+6z)=-5$
$F_z=-\frac{\partial U}{\partial z}=-\frac{\partial }{\partial z}(\frac{3x^2}{2}+5y+6z)=-6$

Thus, the force vector is represented as:
$\overrightarrow{F}=-3x\hat{i}-5\hat{j}-6\hat{k}$

Next, we evaluate the components of the force at $x=6$:
$F_x=-3\left(6\right)=-18\text{N}$
$F_y=-5\text{N}$
$F_z=-6\text{N}$

Now, we find the magnitude of the force vector $\overrightarrow{F}$:
$\left|\overrightarrow{F}\right|=\sqrt{{F_x}^2+{F_y}^2+{F_z}^2}$
$\left|\overrightarrow{F}\right|=\sqrt{{(-18)}^2+{(-5)}^2+{(-6)}^2}$
$\left|\overrightarrow{F}\right|=\sqrt{324+25+36}$
$\left|\overrightarrow{F}\right|=\sqrt{385}\approx 19.62\text{N}$

Rounding 19.62 to the nearest integral value, we obtain 20.