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

Question

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

Accepted Answer

D. 1. When two forces act **perpendicular** to each other on a body, the net (resultant) force is found using the **Pythagorean rule**, since perpendicular vectors combine like the two legs of a right triangle.

Given forces $F_1 = 8 	ext{ N}$ and $F_2 = 6 	ext{ N}$,

$$
F_{net} = \sqrt{F_1^2 + F_2^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10 	ext{ N}
$$

2. By Newton's second law, acceleration equals the net force divided by mass.

Therefore, with $m = 5 	ext{ kg}$,

$$
a = \frac{F_{net}}{m} = \frac{10}{5} = 2 	ext{ m/s}^2
$$

3. To find the **direction** of this resultant relative to the $8$ N force, we use the fact that the angle a resultant makes with one component follows $	an	heta = \dfrac{	ext{opposite side}}{	ext{adjacent side}}$, where the $6$ N force acts perpendicular to the $8$ N force.

So,

$$
	an	heta = \frac{6}{8} = \frac{3}{4}
$$

4. Taking the inverse tangent gives the angle measured from the $8$ N force.

Hence,

$$
	heta = 	an^{-1}\left(\frac{3}{4}ight) 	ext{ from the } 8 	ext{ N force}
$$

Hence, the answer is **Option D**.

The magnitude and direction of the acceleration produced in a body of mass 5 kg5 \text{ kg}5 kg when two mutually perpendicular forces 8 N8 \text{ N}8 N and 6 N6 \text{ N}6 N act on it, are respectively:

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: