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# Answer to Question #113462 in Mechanics | Relativity for Moses

Question #113462
A body moves with velocity v=(5i + 2j - 3k)ms^-1. Under the influence of a constant force F=(4i + 3j - 2k)N. Determine the instantaneous power and the angle between F and V.
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Expert's answer
2020-05-04T13:12:27-0400

Explanation

• Instantaneous power is calculated by F*V.
• When those are given in vector form, power can be found by considering their dot product F.V
• Angle between them is "\\small \\theta"

Calculations

• Magnitude of the uniform velocity

"\\qquad\\qquad\n\\begin{aligned}\n\\small |\\bold{v}| &= \\small \\sqrt{5^2+2^2+(-3)^2}\\\\\n&= \\small \\bold{\\sqrt{38}}\n\\end{aligned}"

• Magnitude of the constant force

"\\qquad\\qquad\n\\begin{aligned}\n\\small |\\bold{F}| &= \\small \\sqrt{4^2+3^2+(-2)^2}\\\\\n&= \\small \\bold{\\sqrt{29}}\n\\end{aligned}"

• Therefore,

"\\qquad\\qquad\n\\begin{aligned}\n\\small \\text{Power} &= \\small \\bold{F.V}\\\\\n&= \\small (4\\bold{i}+3\\bold{j}-2\\bold{k}).(5\\bold{i}+2\\bold{j}-3\\bold{k})\\\\\n&= \\small 20+6+6\\\\\n&= \\small \\bold{32\\,W}\\\\\n\\end{aligned}\n\\\\\n\n\\qquad\\qquad\n\\begin{aligned}\n\\small \\bold{F.V} &= \\small |\\bold{F}|.|\\bold{V}|\\cos\\theta\\\\\n\\small \\cos\\theta &= \\small \\frac{\\bold{F.V}}{|\\bold{F}|.|\\bold{V}|}\\\\\n\\small \\theta &= \\small \\cos^{-1}\\Bigg(\\frac{32}{\\sqrt{29} \\times \\sqrt{38}}\\Bigg)\\\\\n&= \\small \\bold{15.43\\degree}\n\\end{aligned}"

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