A force $$F = (5 + 3y^2)$$ acts on a particle in the $$y$$-direction, where $$F$$ is newton and $$y$$ is in meter. The work done by the force during a displacement from $$y = 2$$ m to $$y = 5$$ m is ______ J.

A force $$F = (5 + 3y^2)$$ N acts on a particle in the $$y$$-direction and the work done as the particle moves from $$y = 2$$ m to $$y = 5$$ m is sought.

Since the force varies with position, the work done is given by the integral $$W = \int_{y_1}^{y_2} F\,dy = \int_2^5 (5 + 3y^2)\,dy$$.

Evaluating the integral gives $$W = \left[5y + 3 \cdot \frac{y^3}{3}\right]_2^5 = \left[5y + y^3\right]_2^5$$.

Substituting the limits into this expression, at $$y = 5$$ we have $$5(5) + 5^3 = 25 + 125 = 150$$, and at $$y = 2$$ we get $$5(2) + 2^3 = 10 + 8 = 18$$. This gives $$W = 150 - 18 = 132$$ J.

The answer is 132.