Consider a force $\overrightarrow{F}=-x \hat{i}+y \hat{j}$. The work done by this force in moving a particle from point $A(1,0)$ to $B(0,1)$ along the line segment is (all quantities are in $SI$ units).

$A$ force of $(6x^2 - 4x + 3) \text{ N}$ acts on a body of mass $0.75 \text{ kg}$ and displaces it from $x = 2 \text{ m}$ to $x = 5 \text{ m}$. The work done by the force is (in $\text{ J}$)

The relation between the displacement $X$ of an object produced by the application of the variable force $F$ is represented by the graph shown in the figure. If the object undergoes a displacement from $X = 0.5 \, m$ to $X = 2.5 \, m$,the work done will be approximately equal to .............. $J$.

$A$ 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?

Power applied to a particle varies with time as $P = (3t^2 - 2t + 1) \text{ W}$. The change in kinetic energy of the particle from $t = 2 \text{ s}$ to $t = 4 \text{ s}$ is ............... $J$.

$A$ 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.