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 $x in [ 0, (13 pi) / 8 ]$ ?

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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 $x in [ 0, (13 pi) / 8 ]$ ?

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Answer 1 · 2016-03-30T17:25:24

$W=(4sin(13/8pi)+13pi)/4 ~~ 2.286$

Explanation:

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:
$W = int_0^(13/8pi) F_x*dx=int_0^(13/8pi) (cosx +2)*dx$
$W=(4sin(13/8pi)+13pi)/4 ~~ 2.286$

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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 $x in [ 0, (13 pi) / 8 ]$ ?

1 Answer

Explanation:

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Science

Math

Humanities

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

1 Answer

Explanation:

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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 $x in [ 0, (13 pi) / 8 ]$ ?