Question

A body of mass 10 kg is acted upon by two perpendicular forces, 6N and 8N. The resultant acceleration of the body is

• A. 1$m{s}^{-2}$ at an angle of $ta{n}^{-1}\left(\frac{3}{4}\right)$w.r.t. 8N force.
• B. 0.2$m{s}^{-2}$ at an angle of $ta{n}^{-1}\left(\frac{3}{4}\right)$w.r.t. 8N force.
• C. 1$m{s}^{-2}$ at an angle of $ta{n}^{-1}\left(\frac{4}{3}\right)$w.r.t. 8N force.
• D. 0.2$m{s}^{-2}$ at an angle of $ta{n}^{-1}\left(\frac{4}{3}\right)$w.r.t. 8N force.

Answer 1

The correct option is A 1$m{s}^{-2}$ at an angle of $ta{n}^{-1}\left(\frac{3}{4}\right)$w.r.t. 8N force.

Here, m=10 kg

The resultant force acting on the body is

$F=\sqrt{(98N{)}^2+(6N{)}^2}$=10N

Let the resultant force F makes an angle $\theta$ w.r.t. 8N force.

From figure, $tan\theta =\frac{6N}{8N}=\frac{3}{4}$

The resultant acceleration of the body is

$a=\frac{F}{m}=\frac{10N}{10kg}=1m{s}^{-2}$

The resulatnt acceleration is along the direction of the resulatnt force.

Hence, the resultant acceleration of the body is 1 $m{s}^{-2}$ at an angle of $ta{n}^{-1}\left(\frac{3}{4}\right)$ w.r.t. 8N force.