Power applied to a particle varies with time as $P = (4t^3 - 5t + 2) \, W$,where $t$ is in seconds. Find the change in its $K.E.$ between time $t = 2 \, s$ and $t = 4 \, s$. ............... $J$

The relationship between force $F$ and displacement $x$ is shown in the figure. The work done by the object for a displacement from $x = 1 \ m$ to $x = 5 \ m$ is equal to ... $J$.

$A$ particle is projected with velocity $v_{0}$ along the $x-$axis. $A$ damping force is acting on the particle which is proportional to the square of the distance from the origin,i.e.,$ma = -\alpha x^{2}$. The distance at which the particle stops is:

$A$ particle of mass $10\,g$ moves in a straight line with retardation $2x$,where $x$ is the displacement in $SI$ units. Its loss of kinetic energy for the above displacement is $\left(\frac{10}{x}\right)^{-n}\,J$. The value of $n$ will be $............$.

$A$ 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 work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20 \, m$ is (in $ \, J$)