A variable force Fx = (14x − 3.0x2) N acts on an object along the x-axis from x1 = -1 m to x2 = +2 m. Calculate the work done by the force.

The work done $W$ by a variable force $F_x$ acting along the x-axis from position $x_1$ to $x_2$ is given by the integral:

$$W = \int_{x_1}^{x_2} F_x\,dx$$

Given $F_x = (14x - 3.0x^2)$ N, $x_1 = -1$ m, and $x_2 = +2$ m.

Substitute the given force function and limits of integration into the formula:

$$W = \int_{-1}^{2} (14x - 3x^2)\,dx$$

Integrate each term with respect to $x$:

$$W = \left[ \frac{14x^{1+1}}{1+1} - \frac{3x^{2+1}}{2+1} \right]_{-1}^{2}$$ $$W = \left[ \frac{14x^2}{2} - \frac{3x^3}{3} \right]_{-1}^{2}$$ $$W = \left[ 7x^2 - x^3 \right]_{-1}^{2}$$

Substitute the upper limit ($x=2$) and subtract the result of substituting the lower limit ($x=-1$):

$$W = (7(2)^2 - (2)^3) - (7(-1)^2 - (-1)^3)$$ $$W = (7(4) - 8) - (7(1) - (-1))$$ $$W = (28 - 8) - (7 + 1)$$ $$W = 20 - 8$$ $$W = 12 \, \text{J}$$

The work done is $+12$ J. Comparing this with the given options, it matches option e.

The final answer is $\text{e. +12 J}$.