Three forces `F_(1)=10 N, F_(2)=8 N, F_(3)=6N` areacting on a particle of mass `5 kg` . The forces `F(2)` and `F(3)` are applied perpendicularly so that particle remains at rest. If the force `F(1)` is removed , then the acceleration of the particle is :

Options

A. `7 ms^(-2)`
B. `2 ms^(-2)`
C. `0.5 ms^(-2)`
D. `4.8 ms^(-2)`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the Forces Acting on the Particle
We have three forces acting on a particle:
- \( F_1 = 10 \, \text{N} \)
- \( F_2 = 8 \, \text{N} \)
- \( F_3 = 6 \, \text{N} \)

Forces \( F_2 \) and \( F_3 \) are applied perpendicularly, and the particle remains at rest under their influence when \( F_1 \) is present.

### Step 2: Calculate the Net Force When \( F_1 \) is Removed
When \( F_1 \) is removed, we need to find the resultant force from \( F_2 \) and \( F_3 \). Since they are perpendicular to each other, we can use the Pythagorean theorem to find the net force \( F_{\text{net}} \):

\[ F_{\text{net}} = \sqrt{F_2^2 + F_3^2} \]

Substituting the values:

\[ F_{\text{net}} = \sqrt{(8 \, \text{N})^2 + (6 \, \text{N})^2} \]

Calculating the squares:

\[ F_{\text{net}} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \text{N} \]

### Step 3: Calculate the Acceleration of the Particle
Now that we have the net force acting on the particle after \( F_1 \) is removed, we can calculate the acceleration \( a \) using Newton's second law:

\[ F = m \cdot a \quad \Rightarrow \quad a = \frac{F_{\text{net}}}{m} \]

Where:
- \( F_{\text{net}} = 10 \, \text{N} \)
- \( m = 5 \, \text{kg} \)

Substituting the values:

\[ a = \frac{10 \, \text{N}}{5 \, \text{kg}} = 2 \, \text{m/s}^2 \]

### Conclusion
The acceleration of the particle after removing \( F_1 \) is \( 2 \, \text{m/s}^2 \).

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