First slide

Simple hormonic motion

Question

What is the minimum coefficient of static friction $μ_{s}$ for which the 1kg block will not slip relative to the lower block if the lower block is displaced by 50 mm to the right from equilibrium position and released ? Take $g = 10 m / s^{2}$ .

Difficult

A

0.346
B

0.628
C

0.525
D

0.428

Solution

$a_{\max} = ω^{2} A and ω^{2} = \frac{k}{(M + m)} w . r . to 6 kg μ_{s} mg \geq {mω}^{2} A \Rightarrow μ_{s} mg \geq m (\frac{k}{M + m}) A \Rightarrow μ_{s} \geq \frac{kA}{(M + m) g} ∴ {(μ_{s})}_{\min} = \frac{600 \times 5 \times 10^{- 2}}{7 \times 10} = 0 .428$

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