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A block of mass m equal to 1 kg placed on top of another block of mass m cost of 5 kg is attached to a horizontal spring of force constant k cost to 20 Newton per metre the coefficient of friction between the blocks Miu friction where is the lower block slides on a frictionless surface the amplitude of the oscillation is 0.4 metre what is the minimum value of miu friction such that the upper block does not sleep over the lower block. Can anyone answer this question?
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A block of mass m equal to 1 kg placed on top of another block of mass...
Understanding the Problem
To determine the minimum coefficient of friction (Miu) required to prevent the upper block from slipping on the lower block, we need to analyze the forces acting on the blocks during oscillation.
Key Parameters
- Mass of the upper block (m1) = 1 kg
- Mass of the lower block (m2) = 5 kg
- Spring constant (k) = 20 N/m
- Amplitude of oscillation (A) = 0.4 m
Forces During Oscillation
1. Maximum Acceleration (a_max)
The maximum acceleration of the system occurs at maximum displacement (A).
Formula: a_max = (k/m_total) * A
Where m_total = m1 + m2 = 6 kg
Thus, a_max = (20 N/m / 6 kg) * 0.4 m = 1.33 m/s².
2. Frictional Force
The maximum static frictional force (F_friction) that can act on the upper block before slipping occurs is given by:
F_friction = Miu * m1 * g, where g = 9.81 m/s².
3. Inertial Force
The inertial force (F_inertial) acting on the upper block due to its acceleration is:
F_inertial = m1 * a_max = 1 kg * 1.33 m/s² = 1.33 N.
Setting Up the Inequality
To prevent slipping, the frictional force must be greater than or equal to the inertial force:
Miu * m1 * g >= F_inertial.
Substituting values:
Miu * 1 kg * 9.81 m/s² >= 1.33 N.
Calculating Miu
Rearranging the inequality to find Miu:
Miu >= 1.33 N / (1 kg * 9.81 m/s²) = 0.135.
Conclusion
Thus, the minimum value of the coefficient of friction (Miu) required to ensure that the upper block does not slip over the lower block is approximately 0.135.
To determine the minimum coefficient of friction (Miu) required to prevent the upper block from slipping on the lower block, we need to analyze the forces acting on the blocks during oscillation.
Key Parameters
- Mass of the upper block (m1) = 1 kg
- Mass of the lower block (m2) = 5 kg
- Spring constant (k) = 20 N/m
- Amplitude of oscillation (A) = 0.4 m
Forces During Oscillation
1. Maximum Acceleration (a_max)
The maximum acceleration of the system occurs at maximum displacement (A).
Formula: a_max = (k/m_total) * A
Where m_total = m1 + m2 = 6 kg
Thus, a_max = (20 N/m / 6 kg) * 0.4 m = 1.33 m/s².
2. Frictional Force
The maximum static frictional force (F_friction) that can act on the upper block before slipping occurs is given by:
F_friction = Miu * m1 * g, where g = 9.81 m/s².
3. Inertial Force
The inertial force (F_inertial) acting on the upper block due to its acceleration is:
F_inertial = m1 * a_max = 1 kg * 1.33 m/s² = 1.33 N.
Setting Up the Inequality
To prevent slipping, the frictional force must be greater than or equal to the inertial force:
Miu * m1 * g >= F_inertial.
Substituting values:
Miu * 1 kg * 9.81 m/s² >= 1.33 N.
Calculating Miu
Rearranging the inequality to find Miu:
Miu >= 1.33 N / (1 kg * 9.81 m/s²) = 0.135.
Conclusion
Thus, the minimum value of the coefficient of friction (Miu) required to ensure that the upper block does not slip over the lower block is approximately 0.135.
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A block of mass m equal to 1 kg placed on top of another block of mass m cost of 5 kg is attached to a horizontal spring of force constant k cost to 20 Newton per metre the coefficient of friction between the blocks Miu friction where is the lower block slides on a frictionless surface the amplitude of the oscillation is 0.4 metre what is the minimum value of miu friction such that the upper block does not sleep over the lower block. Can anyone answer this question?
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