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A constant force = 2.00 n + 3.00 n acts on a 5.00 kg object as it moves in a straight line from the position1 = 1.00 m + 1.00 m to a position 2 = 4.00 m - 1.00 m . determine the work done by the force during this motion.

Respuesta :

force acting on the object is given as

[tex]F = 2.00\hat i + 3.00 \hat j[/tex]

initial and final positions are given as

[tex]r_i = 1.00 \hat i + 1.00\hat j[/tex]

[tex]r_f = 4.00 \hat i - 1.00\hat j[/tex]

now the displacement is given as

[tex]d = r_f - r_i[/tex]

[tex]d = (4.00 - 1.00)\hat i + (-1.00 - 1.00)\hat j[/tex]

[tex]d = 3 \hat i - 2\hat j[/tex]

now the work done is given as

[tex]W = F.d[/tex]

[tex]W = (2.00\hat i + 3.00\hat j).(3\hat i - 2\hat j)[/tex]

[tex]W = 0[/tex]

so there is no work done

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