$A$ position-dependent force $F = 7 - 2x + 3x^2 \, N$ acts on a small body of mass $2 \, kg$ and displaces it from $x = 0$ to $x = 5 \, m$. The work done in joules is

$A$ force $\overrightarrow F = - K(y\hat i + x\hat j)$ (where $K$ is a positive constant) acts on a particle moving in the $x-y$ plane. Starting from the origin,the particle is taken along the positive $x$-axis to the point $(a, 0)$ and then parallel to the $y$-axis to the point $(a, a)$. The total work done by the force $\overrightarrow F$ on the particle is

If a position-dependent force $F(x) = (3x^2 - 2x + 7) \text{ N}$ acting on a body of mass $2 \text{ kg}$ displaces it from $x = 0 \text{ m}$ to $x = 5 \text{ m}$,then the work done by the force is: (in $J$)

The graphs below show the magnitude of the force on a particle as the particle moves along the positive $x$ axis from the origin to $x = x_1$. The force is parallel to the $x$ axis and is conservative. The maximum magnitude $F_1$ has the same value for all graphs. Rank the situations according to the change in the potential energy associated with the force,from least (or most negative) to greatest (or most positive).

$A$ force $F = (5 + 3y^2)$ acts on a particle in the $y$ direction,where $F$ is in newtons and $y$ is in meters. The work done by the force during a displacement from $y = 2 \, m$ to $y = 5 \, m$ is $............. \, J$.

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 $\,J$)