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

Options

A. `1 m s^(-2)` at an angle of `tan^(-1) ((3)/(4))` w.r.t. 6 N force
B. `0.2 m s^(-2) ` at an angle of `tan^(-1)((4)/(3))` w.r.t. 6 N force
C. `1 m s^(-2)` at an angle of `tan^(-1) ((3)/(4))` w.r.t. 8 N force
D. 0.2 `ms^(-2)` at an angle of `tan^(-1)((3)/(4))` w.r.t. 8 N force

Explanation

To find the resultant acceleration of a body acted upon by two perpendicular forces, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the body, \( m = 10 \, \text{kg} \)
- Force 1, \( F_1 = 6 \, \text{N} \)
- Force 2, \( F_2 = 8 \, \text{N} \)

2. **Calculate the Resultant Force:**
Since the forces are perpendicular to each other, we can use the Pythagorean theorem to find the resultant force \( F_{\text{net}} \):
\[ F_{\text{net}} = \sqrt{F_1^2 + F_2^2} = \sqrt{6^2 + 8^2} \]
\[ F_{\text{net}} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{N} \]

3. **Calculate the Acceleration:**
Using Newton's second law of motion, \( F = ma \), we can find the acceleration \( a \):
\[ a = \frac{F_{\text{net}}}{m} = \frac{10 \, \text{N}}{10 \, \text{kg}} = 1 \, \text{m/s}^2 \]

4. **Determine the Direction of the Resultant Force:**
To find the direction of the resultant force, we can use the tangent function:
\[ \tan \theta = \frac{F_2}{F_1} = \frac{8}{6} = \frac{4}{3} \]
Therefore,
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

### Final Result:
The resultant acceleration of the body is:
\[ \boxed{1 \, \text{m/s}^2} \]
at an angle of \( \tan^{-1}\left(\frac{4}{3}\right) \) with respect to the 6 N force.

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