A position dependent force F = (7 – 2x + 3x²)N acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5m. The work done is(in Joule)

Work performed for a small displacement is characterized as:

dW=Fdx_____(1)
the equation's integration
Equation (1) with constraints x=0 to x=5m can be integrated to determine the total effort completed.

$\int dW={\int }_0^{5}Fdx$

$\int dW={\int }_0^5(7-2x+3x^2)dx$

$W={{\left[7x-\frac{2x^2}{2}+\frac{3x^3}{3}\right]}^5}_0$

$W=135J$

Hence the correct answer is 135.