The force-displacement graph for a particle is given. Calculate the work done by the force as the particle moves from $x = 1 \ cm$ to $x = 5 \ cm$. (in $erg$)

$A$ block of mass $1 \ kg$,moving along the $x$-axis with an initial speed $v_{i} = 10 \ m/s$,enters a rough region ranging from $x = 0.1 \ m$ to $x = 1.9 \ m$. The retarding force acting on the block in this range is $F_{r} = -kx \ N$,where $k = 10 \ N/m$. Find the final speed of the block as it crosses the rough region.

The force $F$ acting on a body varies with its position $x$ as shown in the figure. At which point is the body in unstable equilibrium?

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 $\text{ J}$)

$A$ 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.

The force on a particle as a function of displacement $x$ (in $x$-direction) is given by $F = 10 + 0.5x$. The work done corresponding to the displacement of the particle from $x = 0$ to $x = 2$ units is: