$A$ position-dependent force $F = (7 - 2x + 3x^2) \ N$ acts on a small object. The object is displaced from $x = 0$ to $x = 5 \ m$. What is the work done in Joules?

$A$ block of mass $m=1 \; kg$ moving on a horizontal surface with speed $v_{i}=2 \; m \; s^{-1}$ enters a rough patch ranging from $x=0.10 \; m$ to $x=2.01 \; m$. The retarding force $F$ on the block in this range is inversely proportional to $x$ over this range,$F_{r} = -k/x$ for $0.1 < x < 2.01 \; m$,and $F_{r} = 0$ for $x < 0.1 \; m$ and $x > 2.01 \; m$,where $k=0.5 \; J$. What is the final kinetic energy $K_{f}$ and speed $v_{f}$ of the block as it crosses this patch?

The slope of the kinetic energy-displacement curve of a particle in motion is

When a force $\vec{F} = (17 - 2x + 6x^2) \text{ N}$ acts on a body of mass $2 \text{ kg}$ and displaces it from $x = 0 \text{ m}$ to $x = 8 \text{ m}$,the work done is: (in $\text{ J}$)

$A$ body of $0.5 \,kg$ moves along the positive $X$-axis under the influence of a varying force $F$ (in newton) as shown below. If the speed of the object at $x=4 \,m$ is $3.16 \,ms^{-1}$,then its speed at $x=8 \,m$ is ................. $\,ms^{-1}$.

The work done by a force $\vec{F} = -5x^4 \hat{i} \text{ N}$ in displacing a body from $x = 2 \text{ m}$ to $x = -2 \text{ m}$ is: (in $\text{ J}$)