Question

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

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

Answer 1

Answer: (A)

Here, $m$ $=10kg$
The resultant force acting on the body is
$F=\sqrt{{\left(8N\right)}^2+{\left(6N\right)}^2}$ $=10N$
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 resultant acceleration is along the direction of the resultant force.
Hence, the resultant acceleration of the body is $1m$ $s^{-2}$ at an angle of $ta{n}^{-1}$ $\left(\frac{3}{4}\right)$ w.r.t. $8N$ force.