A particle of mass 0.3 kg is subjected to a forces F= - kx with k = 15 N/m. What will be its initial acceleration if it is released from a point x = 20 cm?

To determine the initial acceleration of the particle when it is released from a position of 20 cm, we can start by applying Newton's second law of motion and the given force equation. The force acting on the particle is described by Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position, but in the opposite direction.

Understanding the Force Acting on the Particle

The force acting on the particle is given by:

F = -kx

Here, k is the spring constant (15 N/m), and x is the displacement from the equilibrium position. In this case, the particle is released from x = 20 cm, which we need to convert to meters for consistency in SI units:

• 20 cm = 0.2 m

Calculating the Force

Now, substituting the values into the force equation:

F = - (15 N/m) * (0.2 m)

Calculating this gives:

F = -3 N

The negative sign indicates that the force acts in the opposite direction of the displacement, which is expected for a restoring force.

Applying Newton's Second Law

Next, we apply Newton's second law, which states:

F = ma

Where m is the mass of the particle (0.3 kg) and a is the acceleration we want to find. Rearranging this equation to solve for acceleration gives:

a = F/m

Substituting the Values

Now, substituting the force we calculated and the mass of the particle:

a = -3 N / 0.3 kg

Calculating this results in:

a = -10 m/s²

Interpreting the Result

The negative sign in the acceleration indicates that the particle is accelerating in the direction opposite to its displacement, which is consistent with the behavior of a spring system. Thus, when the particle is released from 20 cm, it will experience an initial acceleration of 10 m/s² directed towards the equilibrium position.

In summary, the initial acceleration of the particle when released from a displacement of 20 cm is -10 m/s², indicating it will move back towards the equilibrium position due to the restoring force exerted by the spring. This behavior is characteristic of simple harmonic motion, where the force and acceleration are always directed towards the equilibrium position.