$A$ horizontal force $F = (g - x^2) \hat{i} \text{ N}$ acts on a wooden block resting on a horizontal smooth surface. The work done to move the block from $x = 0$ to $x = 3 \text{ m}$ (in $\text{J}$) is (Use $g = 10 \text{ m/s}^2$)

$A$ bullet of mass $1 \ kg$ fired with a speed $2 \ m \ s^{-1}$ from $x = 0$ passes through a block of wood whose centre is kept at a distance of $10 \ m$ from the origin as shown in figure. The retarding force $F_r$ on the bullet within the wooden block is $-0.5/x$. The minimum length of the block (up to $1$ decimal digit) required to completely stop the bullet is (Assume $e^4 = 55$). (in $m$)

$A$ force $\vec{F} = (7 - 2x + 3x^2) \, N$ is applied to an object. Calculate the work done as the object moves from $x = 0$ to $x = 5 \, m$. (in $, J$)

$A$ $10 \, kg$ mass moves along the $x-$axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from $x = 0$ to $x = 8 \, cm$?

If a position-dependent force $F(x) = (3x^2 - 2x + 7) \text{ N}$ acting on a body of mass $2 \text{ kg}$ displaces it from $x = 0 \text{ m}$ to $x = 5 \text{ m}$,then the work done by the force is: (in $J$)

$A$ force acts on a particle moving along the $x$-axis,and its variation with position is shown in the figure. At which point will the particle be in a stable equilibrium position?