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mass weighing 10-lb is attached to a spring and stretches a string 6 inches. the medium offers a dampening force that is equal to 5 times its instantaneous velocity. mass released from a point 3 inches below equilibrium with upward velocity 6ft/s. find mass m and spring constant k.

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he mass of the system is approximately 0.63 kg and the spring constant is approximately 44.51 N/m.

What is the mass m and spring constant k?

Given

Mass: 10 lb ≈ 4.54 kg

String stretch: 6 inches ≈ 0.1524 m

Velocity: 6 ft/s ≈ 1.83 m/s

According to Newton's second law, the sum of forces acting on the mass equals its mass times its acceleration:

m * a = k * x - 5v - mg

where a is the acceleration of the mass.

At the moment of release:

Displacement (x) = 0.0762 m (3 inches below equilibrium)

Velocity (v) = 1.83 m/s (upward)

Acceleration (a) is unknown

When the mass initially stretches the string and comes to rest at equilibrium (x = 0 and v = 0), the equation of motion simplifies to:

0 = k * 0 - 5 * 0 - mg

Therefore, k = mg = 4.54 kg * 9.81 m/s² ≈ 44.51 N/m

m * a = 44.51 N/m * 0.0762 m - 5 * 1.83 m/s - 4.54 kg * 9.81 m/s²

a ≈ -14.64 m/s² (downward)

4.54 kg * (-14.64 m/s²) = 44.51 N/m * 0.0762 m - 5 * v - 4.54 kg * 9.81 m/s²

v ≈ 2.96 m/s (upward)

4.54 kg = 44.51 N/m * 0.0762 m / (-14.64 m/s²) + 5 * 2.96 m/s

m ≈ 0.63 kg

Therefore, the mass of the system is approximately 0.63 kg and the spring constant is approximately 44.51 N/m.

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