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A mass of 0.601 kg hangs on a vertical spring of spring constant 19.2 N/m. If...

A mass of 0.601 kg hangs on a vertical spring of spring constant 19.2 N/m. If you displace the mass by 1.69 cm from its equilibrium position and let it go from rest, what will be the maximum velocity?

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Answer #1

Given,

Mass, m = 0.601 Kg

Spring constant, k = 5.83 N/m

The displacement, x = 1.69 cm = 0.0169 m

By the conservation of energy,

1/2 * kx2 = 1/2 * m (vmax)2

kx2 = m (vmax)2

5.83 * (0.0169)2 = 0.601 * (vmax)2

(vmax)2 = 0.0028

vmax = 0.053 m/s

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