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I don't know if you meant
[tex]\vec F = (-3\,\vec\imath + 4\,\vec\jmath - 2\,\vec k)\,\mathrm N[/tex]
or
[tex]\vec F = (-31\,\vec\imath + 43\,\vec\jmath - 2\,\vec k)\,\mathrm N[/tex]
I'll assume the first force is correct.
The object in question undergoes a total displacement of
[tex]\vec d = (-4\,\vec\imath + 4\,\vec\jmath+4\,\vec k)\,\mathrm m - (4\,\vec\imath + \vec\jmath - 5\,\vec k)\,\mathrm m = (-8\,\vec\imath + 3\,\vec\jmath + 9\,\vec k)\,\mathrm m[/tex]
Then the work W done by [tex]\vec F[/tex] along this displacement is
[tex]W = \vec F \cdot \vec d = ((-3)\times(-8)+4\times3+(-2)\times9)=18\,\mathrm{Nm} = \boxed{18\,\mathrm J}[/tex]
Another approach using calculus (it's overkill since [tex]\vec F[/tex] is constant, but it doesn't hurt to check our answer): parameterize the line segment by
[tex]\vec r(t) = (1 - t)(4\,\vec\imath+\vec\jmath-5\,\vec k)\,\mathrm m + t(-4\,\vec\imath+4\,\vec\jmath + 4\,\vec k)\,\mathrm m \\\\ \vec r(t) = \left((4-8t)\,\vec\imath+(1+3t)\,\vec\jmath+(-5+9t)\,\vec k\right)\,\mathrm m[/tex]
with 0 ≤ t ≤ 1.
Then the work W done by [tex]\vec F[/tex] along the given path is equal to the line integral,
[tex]\displaystyle W = \int_0^1 \vec F(\vec r(t)) \cdot \frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt \\\\ W = \int_0^1 \left((-3\,\vec\imath+4\,\vec\jmath-2\,\vec k)\,\mathrm N\right) \cdot \left((-8\,\vec\imath+3\,\vec\jmath+9\,\vec k)\,\mathrm m\right) \,\mathrm dt \\\\ W = \int_0^1((-3)\times(-8)+4\times3+(-2)\times9)\,\mathrm{Nm}\,\mathrm dt \\\\ W = 18\,\mathrm{Nm} \int_0^1\mathrm dt \\\\ W = 18\,\mathrm{Nm} = \boxed{18\,\mathrm J}[/tex]