A particle of mass 0.3 kg subject to a force `F=-kx` with `k=15N//m`. What will be its initial acceleration if it is released from a point 20cm away from the origin?

Options

A. `5 m//s^(2)`
B. `10 m//s^(2)`
C. `3 m//s^(2)`
D. `15 m//s^(2)`

Explanation

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

### Step 1: Identify the given values
- Mass of the particle, \( m = 0.3 \, \text{kg} \)
- Spring constant, \( k = 15 \, \text{N/m} \)
- Initial displacement from the origin, \( x = 20 \, \text{cm} = 0.2 \, \text{m} \)

### Step 2: Calculate the force acting on the particle
The force acting on the particle is given by Hooke's law:
\[ F = -kx \]
Substituting the values:
\[ F = -15 \, \text{N/m} \times 0.2 \, \text{m} = -3 \, \text{N} \]
The negative sign indicates that the force is directed towards the origin.

### Step 3: Calculate the magnitude of the force
Since we are interested in the magnitude of the force:
\[ |F| = 3 \, \text{N} \]

### Step 4: Use Newton's second law to find acceleration
According to Newton's second law:
\[ F = ma \]
We can rearrange this to find the acceleration \( a \):
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{3 \, \text{N}}{0.3 \, \text{kg}} = 10 \, \text{m/s}^2 \]

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

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