Q14 (2024)
The magnitude and direction of the acceleration produced in a body of mass 5 kg when two mutually perpendicular forces 8 N and 6 N act on it, are respectively:
- $2 \, \mathrm{m \, s^{-2}}; \tan^{-1}(3/4) \text{ with } 6\mathrm{ N force}$
- $2 \mathrm{~m \text{ s}^{-2}}; \tan^{-1}(4 / 3) \text{ with } 8\mathrm{ N force}$
- $2 \, \mathrm{m \, s^{-2}}; \tan^{-1}(3/4) \text{ with } 8\mathrm{ N force}$
- $20 \mathrm{~m s}^{-2}; \tan^{-1}(4 / 3) \text{ with } 8\mathrm{ N force}$
### Related Formula
$$F_{\text{net}} = \sqrt{F_1^2 + F_2^2}$$
$$a = \frac{F_{\text{net}}}{m}$$
### Core Logic
Net force $F_{\text{net}} = \sqrt{6^2 + 8^2} = 10\mathrm{N}$.
{{SOL_IMG1}}
Acceleration $a = \frac{10}{5} = 2 \mathrm{m/s}^2$.
Angle with 8 N force: $\tan \theta = \frac{6}{8} \Rightarrow \theta = \tan^{-1}(3/4)$ with 8 N force.
### Pattern Recognition
Perpendicular force resultant vector acceleration.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
