A 5 kg block is placed on an inclined plane with an angle of 30 degrees. If the coefficient of kinetic friction between the block and the plane is 0.2, calculate the acceleration of the block when it's released.
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To calculate the acceleration of the block when it's released, we need to consider the forces acting on it.
The weight force (mg) can be resolved into two components:
- The force acting parallel to the inclined plane (mg*sin θ)
- The normal force perpendicular to the inclined plane (mg*cos θ)
The force of kinetic friction (fk) can be calculated using the equation:
fk = μ * N
where μ is the coefficient of kinetic friction and N is the normal force.
The net force acting parallel to the inclined plane (Fnet) can be calculated as the difference between the force parallel to the inclined plane and the force of kinetic friction:
Fnet = m * g * sin θ - fk
To calculate the acceleration (a), we can use Newton's second law:
Fnet = m * a
Substituting the values into the equations:
Fnet = (5 kg) * (9.8 m/s^2) * sin 30° - (0.2) * (5 kg) * (9.8 m/s^2) * cos 30°
= (5 kg) * (4.9 m/s^2) - (0.2) * (5 kg) * (9.8 m/s^2) * (0.866)
≈ 24.5 N - 8.487 N
≈ 16.013 N
Fnet = m * a
16.013 N = (5 kg) * a
Solving for acceleration:
a = 16.013 N / 5 kg
≈ 3.2026 m/s^2
Therefore, the acceleration of the block when it's released is approximately 3.2026 m/s^2.
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