Question

A body of mass 5 kg is acted upon by two perpendicular forces 8 n and 6n. Give the magnitude and direction of the acceleration of the body.

Answer 1

consider forces in their vector form and now perform vector addition using this and find the anagle between resultant vector and original vector using triangle law of vector addition
$\sqrt { {a}^{2} + {b}^{2} + 2abcos(90) }$

Answer 2

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$\bold\red{hello...frd\:swigy\:here}$

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Mass of the body, m = 5 kg

The given situation can be represent as shown in attachment (please see attached file)

The resultant of two forces is given as:

R = √{(8)^2 + (-6)^2}=  √(64+36) = 10 N

θ is the angle made by R with the force of 8 N

Θ = tan^-1(-6/8) = -36.87°

The negative sign indicates that θ is in the clockwise direction with respect to the force of magnitude 8 N.

As per Newton’s second law of motion, the acceleration (a) of the body is given as:

F = ma

=> a= F/m = 10/5 = 2 m/s^2

I hope, this will help you

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