Answer to Question #113462 in Mechanics | Relativity for Moses

Question

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.

Answer 1

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

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

• Magnitude of the constant force

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

• Therefore,

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