If extension in the spring at equilibrium position is 6 cm then acceleration of block if it is released by further streching 4 cm.

To determine the acceleration of the block when it is released after being stretched an additional 4 cm beyond its equilibrium position, we can use Hooke's Law and Newton's second law of motion. Let's break this down step by step.

Understanding the System

In this scenario, we have a spring that is initially at its equilibrium position when it is stretched by 6 cm. When the block is further stretched by 4 cm, the total extension becomes 10 cm. The spring force will act to restore the block back to its equilibrium position.

Applying Hooke's Law

Hooke's Law states that the force exerted by a spring is proportional to its extension or compression from the equilibrium position. Mathematically, it is expressed as:

F = -kx

Here, F is the restoring force, k is the spring constant, and x is the displacement from the equilibrium position.

Calculating the Force

In our case, when the spring is stretched an additional 4 cm (or 0.04 m), the total displacement x from the equilibrium position is:

• Initial extension = 6 cm = 0.06 m
• Additional stretch = 4 cm = 0.04 m
• Total extension, x = 0.06 m + 0.04 m = 0.10 m

Thus, the force exerted by the spring when stretched to 10 cm is:

F = -k(0.10)

Finding the Acceleration

According to Newton's second law, the acceleration a of the block can be found using the formula:

F = ma

Where m is the mass of the block. Rearranging this gives us:

a = F/m

Substituting the force from Hooke's Law into this equation, we get:

a = -kx/m

Final Steps

To find the exact value of acceleration, we need the spring constant k and the mass m of the block. However, the direction of the acceleration will be towards the equilibrium position, meaning it will be negative in this context since it opposes the direction of the stretch.

In summary, the acceleration of the block when released after being stretched an additional 4 cm can be calculated using the spring constant and the mass of the block. The key takeaway is that the block will accelerate back towards the equilibrium position due to the restoring force of the spring.