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A 0.7 kg mass is attached to an ideal spring with a constant of 86 N/m. The mass is initially held at rest so that the spring is at its unextended length of 0.95 m. The mass is then released. What is the maximum distance the mass will fall?

User Alcsan
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Since the mass is attached to an ideal spring, the system will undergo simple harmonic motion. The maximum distance the mass will fall is equal to the amplitude of the oscillation.

The period of oscillation can be calculated as:

T = 2π√(m/k)

where m is the mass and k is the spring constant.

Substituting the given values, we get:

T = 2π√(0.7 kg / 86 N/m) ≈ 0.887 s

The maximum distance the mass will fall is equal to half the amplitude of the oscillation, which can be calculated using the equation:

x = A cos(2πt/T)

where x is the displacement of the mass from its equilibrium position at time t, and A is the amplitude of oscillation.

At the maximum displacement, cos(2πt/T) will be equal to -1. Therefore,

A = -x

The velocity of the mass at the maximum displacement will be zero. Therefore, the total energy of the system will be equal to the potential energy at the maximum displacement:

1/2 k A^2 = m g A

where g is the acceleration due to gravity.

Solving for A, we get:

A = (m g / k) = (0.7 kg x 9.81 m/s^2) / 86 N/m ≈ 0.0807 m

Therefore, the maximum distance the mass will fall is approximately 0.0807 m.

User Lovlesh
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