The magnitude and direction of the accelerati | vedclass

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Question

(A) Given mass $m = 5 	ext{ kg}$.The two forces are $F_1 = 8 	ext{ N}$ and $F_2 = 6 	ext{ N}$,which are mutually perpendicular.The resultant force

Vedclass · Accepted Answer

$2 	ext{ ms}^{-2}$; $	an^{-1}(3/4 	ext{ with } 8 	ext{ N force})$

Vedclass · Answer

(A) Given mass $m = 5 	ext{ kg}$.The two forces are $F_1 = 8 	ext{ N}$ and $F_2 = 6 	ext{ N}$,which are mutually perpendicular.The resultant force $F$ is given by $F = \sqrt{F_1^2 + F_2^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 	ext{ N}$.Using Newton's second law,the acceleration $a = F/m = 10 	ext{ N} / 5 	ext{ kg} = 2 	ext{ ms}^{-2}$.The direction $	heta$ of the resultant force with respect to the $8 	ext{ N}$ force is given by $	an 	heta = \frac{F_2}{F_1} = \frac{6}{8} = 3/4$.Therefore,$	heta = 	an^{-1}(3/4)$ with the $8 	ext{ N}$ force.Thus,the correct option is $A$.

The magnitude and direction of the acceleration produced in a body of mass $5 \text{ kg}$ when two mutually perpendicular forces $8 \text{ N}$ and $6 \text{ N}$ act on it,are respectively:

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