Question

A constant force F = -31+43 - 2k is applied to an object that is moving along a straight line from the point (4,1,-5) to the point (-4,4,4). Find the work done if the distance is measured in meters and the force in newtons. Include units in your answer. (Note, units are case sensative. Clicking on the link units will give a list of units.)

Answer 1

I don't know if you meant

$\vec F = (-3\,\vec\imath + 4\,\vec\jmath - 2\,\vec k)\,\mathrm N$

or

$\vec F = (-31\,\vec\imath + 43\,\vec\jmath - 2\,\vec k)\,\mathrm N$

I'll assume the first force is correct.

The object in question undergoes a total displacement of

$\vec d = (-4\,\vec\imath + 4\,\vec\jmath+4\,\vec k)\,\mathrm m - (4\,\vec\imath + \vec\jmath - 5\,\vec k)\,\mathrm m = (-8\,\vec\imath + 3\,\vec\jmath + 9\,\vec k)\,\mathrm m$

Then the work W done by $\vec F$ along this displacement is

$W = \vec F \cdot \vec d = ((-3)\times(-8)+4\times3+(-2)\times9)=18\,\mathrm{Nm} = \boxed{18\,\mathrm J}$

Another approach using calculus (it's overkill since $\vec F$ is constant, but it doesn't hurt to check our answer): parameterize the line segment by

$\vec r(t) = (1 - t)(4\,\vec\imath+\vec\jmath-5\,\vec k)\,\mathrm m + t(-4\,\vec\imath+4\,\vec\jmath + 4\,\vec k)\,\mathrm m \\\\ \vec r(t) = \left((4-8t)\,\vec\imath+(1+3t)\,\vec\jmath+(-5+9t)\,\vec k\right)\,\mathrm m$

with 0 ≤ t ≤ 1.

Then the work W done by $\vec F$ along the given path is equal to the line integral,

$\displaystyle W = \int_0^1 \vec F(\vec r(t)) \cdot \frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt \\\\ W = \int_0^1 \left((-3\,\vec\imath+4\,\vec\jmath-2\,\vec k)\,\mathrm N\right) \cdot \left((-8\,\vec\imath+3\,\vec\jmath+9\,\vec k)\,\mathrm m\right) \,\mathrm dt \\\\ W = \int_0^1((-3)\times(-8)+4\times3+(-2)\times9)\,\mathrm{Nm}\,\mathrm dt \\\\ W = 18\,\mathrm{Nm} \int_0^1\mathrm dt \\\\ W = 18\,\mathrm{Nm} = \boxed{18\,\mathrm J}$