In the arrangement shown coefficient of friction between the two blocks A and B is 0.6 and ground is frictionless. Force constant of spring is 900 N/m. Find the maximum elongation of spring (in cm) such that the blocks do not slip over each other when released.Correct answer is '1'. Can you explain this answer?

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In the arrangement shown coefficient of friction between the two blocks A and B is 0.6 and ground is frictionless. Force constant of spring is 900 N/m. Find the maximum elongation of spring (in cm) such that the blocks do not slip over each other when released.

Correct answer is '1'. Can you explain this answer?

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In the arrangement shown coefficient of friction between the two block...

Frictional Force Between the Blocks: The frictional force between the two blocks can be calculated using the formula:
f_friction = μ ⋅ N
Where:

μ = 0.6 is the coefficient of friction.
N is the normal force, which is equal to the weight of block A.

For block A (1 kg):
N = mA ⋅ g = 1⋅9.8 = 9.8N
Now, calculate the maximum frictional force:

_friction= 0.6⋅9.8 = 5.88 N
Maximum Acceleration Without Slipping: The maximum force that can be applied to block A without it slipping over block B is the frictional force we calculated. This force will cause an acceleration in both blocks (since they are stacked together).
Using Newton’s second law:

This is the maximum acceleration the system can have without the blocks slipping over each other.
Force on the System Due to Spring: The spring exerts a restoring force given by Hooke’s law:
F_spring = k ⋅ x
Where:

k = 900 N/m is the spring constant.
x is the elongation in the spring.

This spring force will accelerate both blocks together, so the total mass being accelerated is:

Applying Newton’s second law for the whole system:

The maximum elongation of the spring such that the blocks do not slip over each other is approximately 1 cm (rounded to the nearest whole number).

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