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 :

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

Answer 1

Answer: (A)

$1ms^{- 2}$at an angle of $\tan^{- 1}\left( \frac{3}{4} \right)$w.r.t. 8 N force

Answer 2

Here m= 10 kg

The resultant force acting on the body is

$F = \sqrt{(8N)^{2} + (6N)^{2}} = 10N$

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

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

The resultant acceleration is along the direction of the resultant force

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