A force F~ = Fx ˆı + Fy ˆ acts on a particle that
undergoes a displacement of ~s = sx ˆı + sy ˆ
where Fx = 8 N, Fy = −1 N, sx = 6 m, and
sy = 2 m.
Find the work done by the force on the
particle.
Answer in units of J.

Respuesta :

Answer:

W = 46 J

Explanation:

We need to find the angle between the two vectors Force vector and displacement vector.

First we will find the angle α of the force vector

[tex]tan\alpha =\frac{1}{8} \\\\\\alpha =7.125 deg\\[/tex]

Then we find the angle β of the displacement vector

[tex]tan\beta=\frac{2}{6} \\\\beta = 18.43 deg\\[/tex]

With these two angles we can find the angle between the two vectors

∅ = α + β = 25.56 deg

The definition of work is given by the expression

[tex]W=F*d*cos (theta)[/tex]

The absolute value of F will be:

[tex]F=\sqrt{8^{2}+1^{2}  } \\F= 8.06 N[/tex]

The absolute value of d will be:

[tex]d=\sqrt{(6 )^{2}+(2)^{2}  } \\d= 6.32m\\[/tex]

Now we have:

[tex]W=8.06*6.32*cos(25.56)\\W=46 J[/tex]

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