A position-dependent force \[f = 7 - 2x + 3{x^2}\] newton acts on a small body of mass 2 kg and displaces from \[x = 0\] to \[x = 5\]. Calculate the work done by the force.

Hint: The given force acting on the small body is the variable force. Recall the expression for the work done by the variable force. Substitute the limits of integration from \[x = 0\] to \[x = 5\] and evaluate the work done by the given force.

Formula used:
Work done, \[W = \int\limits_{{x_i}}^{{x_f}} {F\,dx} \]
Here, F is the variable force.

Complete step by step answer:
We have given the variable force is acting on the body of mass \[m = 2\,{\text{kg}}\] which displaces its position from \[{x_i} = 0\] to \[{x_f} = 5\] meter. We have the expression for the work done by the variable force,
\[W = \int\limits_{{x_i}}^{{x_f}} {F\,dx} \]
Here, F is the variable force.

Substituting \[f = 7 - 2x + 3{x^2}\], \[{x_i} = 0\] and \[{x_f} = 5\] in the above equation, we get,
\[W = \int\limits_0^5 {\left( {7 - 2x + 3{x^2}} \right)\,dx} \]
\[ \Rightarrow W = \left( {7x} \right)_0^5 - 2\left( {\dfrac{{{x^2}}}{2}} \right)_0^5 + 3\left( {\dfrac{{{x^3}}}{3}} \right)_0^5\]
\[ \Rightarrow W = 35 - 25 + 125\]
\[ \therefore W = 135\,{\text{J}}\]

Therefore, the work done by the variable force f on the small body is 135 J.

Additional information:
If the constant force acts on the body throughout its path, the work done by the constant force is expressed as, \[W = F\,dx\], where, F is the constant force and \[dx\] is the displacement of the body. If the variable force acts on the body, the magnitude of the force changes at every position and the work done by such force is expressed as,
\[W = \int\limits_{{x_i}}^{{x_f}} {F\,dx} \].
As we know the work done does not depend on the path of the motion and it depends only on the initial and final position of the body. If the displacement of the body is zero, the work done will also be the zero.

Note: In the solution, we have assumed that the displacement of the body is in meters since the force and mass are given in S.I units. The S.I unit of the work done is N m and its derived unit is joule. The expression for the work done by the variable force is obtained from the known expression, \[F = \dfrac{{dW}}{{dx}}\]and integrating it to obtain \[W = \int\limits_{{x_i}}^{{x_f}} {F\,dx} \].