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:

The displacement $s$ of a body of mass $3 \,kg$ under the action of a force is given by $s = \frac{t^3}{3}$, where $s$ is in metres and $t$ is in seconds. The work done by the force in the first two seconds is (in $\,J$)

$A$ 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......$J$.

For a particle moving under the action of a variable force,the kinetic energy-position $(K-x)$ graph is given. Then:

The kinetic energy $(E_k)$ of a particle moving along the $X$-axis varies with its position $(X)$ as shown in the figure. The force acting on the particle at $X = 10 \ m$ is

$A$ force $\vec{F} = (7 - 2x + 3x^2) \, N$ is applied to an object. Calculate the work done as the object moves from $x = 0$ to $x = 5 \, m$. (in $, J$)