The force applied against a moving object travelling on a linear path is given by F(x)= cosx + 2 . How much work would it take to move the object over x in [ 0, (13 pi) / 8 ] ?

Mar 30, 2016

$W = \frac{4 \sin \left(\frac{13}{8} \pi\right) + 13 \pi}{4} \approx 2.286$

Explanation:

Given: Force, $F \left(x\right) = \cos \left(x\right) + 2$
Required: Work done over $x \in \left[0 , \frac{13 \pi}{8}\right]$
Solution Strategy: Use the Work/ Force formula:
dW = vecF*vecdr; W= F_x*dx + F_y*dy+F_z*dz= |F_x|*dx
Since $\vec{F} = {F}_{x}$ only then we integrate in $\mathrm{dx}$ only:
$W = {\int}_{0}^{\frac{13}{8} \pi} {F}_{x} \cdot \mathrm{dx} = {\int}_{0}^{\frac{13}{8} \pi} \left(\cos x + 2\right) \cdot \mathrm{dx}$
$W = \frac{4 \sin \left(\frac{13}{8} \pi\right) + 13 \pi}{4} \approx 2.286$