Problem 6.18

6.18. a)

We know that

$W = \int^{f}_{i} \vec{F} \cdot d\vec{s}$

Applying the $\vec{F}$ that we have in this problem, we find that

$W = 15000 x + \frac{10000 x^{2}}{2} - \frac{25000 x^{3}}{3} \Big\vert^{0.6}_{0}$

where we have used the fact that the angle between $\vec{F}$ and $d\vec{s}$ is $0{}^{\circ}$.

Solving for work we have

W = 9 kJ + 1.8 kJ - 1.8 kJ  =  9.00 kJ

b)

Now, we are evaluating our integral from 0 to 1 instead of from zero to 0.6. We get the following.

$W = (15 kN)(1 m) + \frac{(10 kN/m)(1 m)^{2}}{2} - \frac{(25 kN/m^{2})(1 m)^{3}}{3}$

W = 11.7 kJ which is smaller than in part a) by 29.6%.