$2 \,m / s ^{2},$ at an angle of $37^{\circ}$ with a force of $8 \,N$

Mass of the body, $m=5\, kg$

The given situation can be represented as follows

The resultant of two forces is given as

$R=\sqrt{(8)^{2}+(-6)^{2}}=\sqrt{64+36}=10 \,N$

$\theta$ is the angle made by $R$ with the force of $8 \,N$ $\therefore \theta=\tan ^{-1}\left(\frac{-6}{8}\right)=-36.87^{\circ}$

The negative sign indicates that $\theta$ 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:

$F=m a$

$\therefore a=\frac{F}{m}=\frac{10}{5}=2 \,m / s ^{2}$