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A mass m is put atop of a mass M. The mass M vibrates horizontally without friction with the amplitude A = 4 cm at the end of a horizontal Hookean spring. Find the minimal coefficient of friction such that m is not moving with respect to M. Take $k = 1000 {Nm}^{- 1}$ and $m = 2 kg, M = 6 kg .$

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Detailed Solution

Force by spring $K x = (m + M) a$

for x = A = 4 cm, Maximum acceleration $a = μ g$

to move together,

$1000 \times 4 / 100 = (2 + 6) μ \times 10 on$

solving $μ = 0.5$

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