If coefficient of friction between the block and plane inclined at 30 degree is 0.6.then the acceleration of block will be. 0 g/2 g/4 g?

Question

Answer

Answer

Explanation:

Given data:
Coefficient of friction (μ) = 0.6
Angle of inclination (θ) = 30 degrees

Calculating acceleration:
The force of friction (Ff) can be calculated using the formula:
Ff = μ * N
where N is the normal force acting on the block perpendicular to the inclined plane.

The normal force (N) can be resolved into two components:
N = mg * cos(θ)
where m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of inclination.

The gravitational force acting parallel to the inclined plane is:
mg * sin(θ)

The net force acting on the block along the inclined plane is:
Fnet = mg * sin(θ) - Ff
= mg * sin(θ) - μ * mg * cos(θ)

Using Newton's second law of motion:
Fnet = ma
where a is the acceleration of the block.

Substitute the expressions for Fnet and Ff into the equation:
mg * sin(θ) - μ * mg * cos(θ) = ma

Solving for acceleration (a):
a = g * (sin(θ) - μ * cos(θ))

Calculating the acceleration:
Substitute the given values:
θ = 30 degrees
μ = 0.6
a = g * (sin(30) - 0.6 * cos(30))

Calculate the acceleration value to determine if it is 0 g, 2 g, or 4 g.

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