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 ] #?

Question

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  ]  #?

William Audy · Accepted Answer

#W=(4sin(13/8pi)+13pi)/4 ~~ 2.286# Given: Force, #F(x)=cos(x)+2# Required: Work done over #x in [0,(13pi)/8]# Solution Strategy: Use the Work/ Force formula: #dW = vecF*vecdr; W= F_x*dx + F_y*dy+F_z*dz= |F_x|*dx# Since #vecF = F_x# only then we integrate in #dx# only: #W = int_0^(13/8pi) F_x*dx=int_0^(13/8pi) (cosx +2)*dx# #W=(4sin(13/8pi)+13pi)/4 ~~ 2.286#

Amelia Adams · Answer

undefined To find the work done, we need to integrate the force function $ F(x) $ over the given interval $[0, \frac{13\pi}{8}]$ with respect to $ x $:

$$ W = \int_{0}^{\frac{13\pi}{8}} \left( \cos(x) + 2 \right) \, dx $$

$$ W = \left[ \sin(x) + 2x \right]_{0}^{\frac{13\pi}{8}} $$

$$ W = \left( \sin\left(\frac{13\pi}{8}\right) + 2\cdot\frac{13\pi}{8} \right) - \left( \sin(0) + 2\cdot0 \right) $$

$$ W = \left( \sin\left(\frac{13\pi}{8}\right) + \frac{13\pi}{4} \right) - 0 $$

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

$$ W \approx -0.707 + 10.21 $$

$$ W \approx 9.503 $$