If a force of $(6x^2 - 4x) \ N$ acts on a body of mass $10 \ kg$,then the work done by the force in displacing the body from $x = 2 \ m$ to $x = 4 \ m$ is: (in $J$)

$A$ sphere of mass $m$ is set in motion with initial velocity $v_0$ on a surface where $kx^n$ is the frictional force,with $k$ and $n$ as constants and $x$ as the distance from the starting point. Find the distance at which the sphere will stop.

$A$ force $F$ acting on an object varies with distance $x$ as shown in the graph. The force is in newton $(N)$ and $x$ is in metre $(m)$. The work done by the force in moving the object from $x = 0$ to $x = 6\,m$ is......$J$.

$A$ 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 ................. $\,ms^{-1}$.

$A$ block of mass $2\,kg$ moving on a horizontal surface with speed of $4\,ms^{-1}$ enters a rough surface ranging from $x = 0.5\,m$ to $x = 1.5\,m$. The retarding force in this range of rough surface is related to distance by $F = -kx$ where $k = 12\,Nm^{-1}$. The speed of the block as it just crosses the rough surface will be ........... $ms^{-1}$.

$A$ particle is moved along a path $A-B-C-D-E-F-A$,as shown in the figure,in the presence of a force $\vec{F} = (\alpha y \hat{i} + 2 \alpha x \hat{j}) \ N$,where $x$ and $y$ are in meters and $\alpha = -1 \ N/m$. The work done on the particle by this force $\vec{F}$ will be . . . . . . Joule.