Question

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

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

Answer 1

Answer: (A)

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