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3 years ago

10

A straight ramp rises from the ground at an angle of 25.5 degrees and has a base that is 30 feet long. How high is the ramp? Rou

nd to the nearest inch

1 answer:

9966 [12]3 years ago

7 0

The height of the ramp is 14 feet.

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3 years ago

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 poi

patriot [66]

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 <em>W</em> 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 ≤ <em>t</em> ≤ 1.

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

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

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3 years ago