$A$ body of mass $5 \; kg$ is acted upon by two perpendicular forces $8 \; N$ and $6 \; N$. Find the magnitude and direction of the acceleration of the body.

(N/A) The magnitude of the resultant force $R$ is given by the Pythagorean theorem for two perpendicular forces:
$R = \sqrt{(8 \; N)^2 + (6 \; N)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \; N$
According to Newton's second law of motion,the magnitude of acceleration $a$ is:
$a = \frac{F}{m} = \frac{10 \; N}{5 \; kg} = 2 \; m/s^2$
The direction of the acceleration $\theta$ with respect to the $8 \; N$ force is given by:
$\tan \theta = \frac{6 \; N}{8 \; N} = 0.75$
$\theta = \tan^{-1}(0.75) \approx 36.87^{\circ}$
Thus,the acceleration is $2 \; m/s^2$ at an angle of $36.87^{\circ}$ with the $8 \; N$ force.