A force F = ( 5 + 3 y ^ 2 ) acts on a particle in the y -direction,...

If an object is moved by certain distance by the application of an external force along the direction of the force, then the force is said to perform some work.

Given, the force exerted on the object is

$F = 5 + 3{y}^{2}$

The formula to calculate the work done $\left(W\right)$ by the force in moving the object from a position ${y}_{1}$ to another position ${y}_{2}$ can be written as

$W = {∫}_{{y}_{1}}^{{y}_{2}}Fdy . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 )$

Substitute the values into equation (1) and solve to calculate the required work done by the force.

$\begin{cases}W & = & {∫}_{2}^{5}\left(5 + 3{y}^{2}\right)dy \\ = & {\left[5y + \dfrac{3{y}^{3}}{3}\right]}_{2}^{5} \\ = & 132\end{cases}$