A position dependent force F = (7 - 2x + 3x² ) N acts on a small object of mass 2 kg to displace it from x = 0 to x = 5m . The work done in joule is

Correct Option: The correct option is 135 J.

Step-by-step Explanation:

To find the work done by a variable force $F\left(x\right)$ acting on an object as it moves from an initial position $x_i$ to a final position $x_f$, we use the integration of force with respect to displacement:

$W={\int }_{x_i}^{x_f}F\left(x\right) dx$

Here, the force function is given by:
$F\left(x\right)=7-2x+3x^2$

The object is displaced from the initial position $x_i=0$ to the final position $x_f=5\text{m}$.

Substituting the force function and the limits into the work equation, we get:
$W={\int }_0^5(7-2x+3x^2) dx$

Now, perform the integration term-by-term:
$\int 7 dx=7x$
$\int -2x dx=-x^2$
$\int 3x^2 dx=x^3$

Combining the integrated terms, we have:
$W={[7x-x^2+x^3]}_0^5$

Substitute the limits into the integrated expression:
$W=\left(7\right(5)-{\left(5\right)}^2+{\left(5\right)}^3)-\left(7\right(0)-{\left(0\right)}^2+{\left(0\right)}^3)$

Simplify each component:
$W=(35-25+125)-0$
$W=10+125$
$W=135\text{J}$

Thus, the total work done is 135 J.