A partical of mass 0.3 kg is subjected to a force F = - kx , Where = `15 Nm^(-1)` . What will be its initial acceleration when particale is relesed from a point 20 cm away from origin ?

Options

A. `3 ms^(-2)`
B. `15 ms^(-2)`
C. `5 ms^(-2)`
D. `10 ms^(-2)`

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Force Equation
The force acting on the particle is given by Hooke's law:
\[ F = -kx \]
where:
- \( k = 15 \, \text{N/m} \) (spring constant)
- \( x \) is the displacement from the equilibrium position.

### Step 2: Convert Displacement to Meters
The displacement given is 20 cm. We need to convert this to meters:
\[ x = 20 \, \text{cm} = \frac{20}{100} \, \text{m} = 0.2 \, \text{m} \]

### Step 3: Calculate the Force
Now, we can calculate the force using the formula:
\[ F = -kx \]
Substituting the values:
\[ F = -15 \, \text{N/m} \times 0.2 \, \text{m} \]
\[ F = -3 \, \text{N} \]

### Step 4: Use Newton's Second Law to Find Acceleration
According to Newton's second law:
\[ F = ma \]
where:
- \( m = 0.3 \, \text{kg} \) (mass of the particle)
- \( a \) is the acceleration.

Rearranging the equation to solve for acceleration:
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{-3 \, \text{N}}{0.3 \, \text{kg}} \]
\[ a = -10 \, \text{m/s}^2 \]

### Step 5: Conclusion
The initial acceleration of the particle when released from a point 20 cm away from the origin is:
\[ a = -10 \, \text{m/s}^2 \]