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ololo11 [35]
3 years ago
6

A block on a horizontal frictionless plane is attached to a spring, as shown below. The block oscillates along the x-axis with s

imple harmonic motion of amplitude A. m k −A 0 +A vx0 Which statement about the block is correct?a. At x = A, its acceleration is zero.b. At x = 0, its velocity is zero.c. At x = A, its velocity is at a maximum.d. At x = 0, its acceleration is at a maxi
Physics
1 answer:
AleksandrR [38]3 years ago
8 0

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

Answer:

D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

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B and C

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Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1
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Answer:

w = 4,786 rad / s ,  f = 0.76176 Hz

Explanation:

For this problem let's use the concept of angular momentum

       L = I w

The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved

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      L₀ = 0 + I₂ w₂

Final after coupling

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The moments of inertia of a disk with an axis of rotation in its center are

      I = ½ M R²

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Let's reduce the units to the SI System

      d₁ = 60 cm = 0.60 m

      d₂ = 40 cm = 0.40 m

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Angular velocity and frequency are related.

      w₂ = 2 π f₂

      w₂ = 2π 3.33

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Let's replace

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       w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

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      w = 2π f

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