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

Solution & Explanation

1. Calculate the Restoring Force at the Release Point

The particle is subjected to a linear restoring force directed toward the origin, given by the expression:

$$F = -k \cdot x$$

We are given the following values from the problem statement:

• Spring constant / force constant: $$k = 15 \,\, \text{N/m}$$
• Release position displacement from the origin: $$x = 20 \,\, \text{cm}$$

First, we convert the displacement from centimeters into SI units (meters):

$$x = \frac{20}{100} = 0.2 \,\, \text{m}$$

Substituting these parameters into the magnitude of the restoring force equation:

$$|F| = 15 \cdot 0.2 = 3.0 \,\, \text{N}$$

2. Determine the Initial Acceleration using Newton's Second Law

According to Newton's Second Law of Motion, the net force acting on a body is equal to the product of its mass ($$m$$) and its acceleration ($$a$$):

$$F = m \cdot a$$

We isolate the acceleration variable ($$a$$) by dividing force by the mass ($$m = 0.3 \,\, \text{kg}$$):

$$a = \frac{|F|}{m}$$

$$a = \frac{3.0}{0.3} = 10 \,\, \text{m/s}^2$$

Concept Check: Because the force is directly proportional to displacement ($$F \propto x$$), the particle experiences its maximum force and maximum acceleration at the exact instant it is released from its point of furthest displacement ($$x = 20 \,\, \text{cm}$$).

Correct Option Key: $$10 \,\, \text{m/s}^2$$