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**Hint:**The work done by a system is calculated by multiplying the force acting it with the displacement of the system for this force. When there are different types of force acting upon a system we integrate the product of forces and the displacements in between required limits to get the total work done by the system.

First, we have to find the total work done from the integration process with the given equation of the force and thereafter, put the given values of existing constants to find the work done.

**Formula used:**

The total work done, $W = \int_0^x F .dx$

$F$ is the applied force and $x$ is the displacement.

**Complete step by step answer:**

The force $F$ is acting on a particle varies with the displacement $x$ as \[F = ax - b{x^2}\](given)

The work done by a system is the product of applied force and the displacement due to force on the system.

So, The total work done, $W = \int_0^x F .dx$

By putting the equation of the force $F$ given above, we get

\[W = \int_0^x {(ax - b{x^2})} .dx\]

\[ \Rightarrow W = \int_0^x {(axdx - b{x^2}dx)} \]

\[ \Rightarrow W = \left[ {\dfrac{{a{x^2}}}{2} - \dfrac{{b{x^3}}}{3}} \right]_0^x\]

\[ \Rightarrow W = \left[ {\dfrac{{a{x^2}}}{2} - \dfrac{{b{x^3}}}{3}} \right]\]

The values of the constants are given by, \[a = 1N/m\] and \[b = 1N/{m^2}\]

Putting these values we get,

\[ \Rightarrow W = \left[ {\dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3}} \right]\]

So, for displacement $x = 1meter$

The work done will be,

\[ \Rightarrow W = \left[ {\dfrac{1}{2} - \dfrac{1}{3}} \right]\]

$ \Rightarrow W = \dfrac{1}{6}$

We know the S.I unit of work done is Joule. So, the work done is $\dfrac{1}{6}J$

**So, the correct answer is “Option A”.**

**Note:**Work, in physics, amount of energy transfer that happens once the object is displaced over a distance by an external force at a minimum of a part that is applied along the direction of the displacement. If the force is constant, work could also be computed by multiplying the length of the trail by the element of the force acting on the trail. To show this idea mathematically, the work W is adequate to the force of times the displacement $x$, or\[W = Fx\] . If the force is being exerted at an angle \[\theta \] to the displacement, the work done is \[W = Fx\cos \theta \] .

Work done on a body is accomplished not solely by a displacement of the body as a full from one place to a different however additionally, for instance, by pressure a gas, by rotating a shaft, and even by inflicting invisible motions of the particles among a body by an external magnetism.

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