# Minimum mass required

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For mas m to move
$mg \sin 37°+ \mu mg\cos 37°=kx =(6/5)mg$
$kx=Mg$
hence M should be equal to (6/5)m

1 vote

You are assuming that the maximum tension in the spring is $Mg$ and occurs when the mass $M$ hangs stationary. The maximum tension in the spring is $2Mg$ and occurs when this mass has fallen and the spring has reached maximum extension before mass $M$ moves back up again.
One aspect which worries me is that the condition for no sliding by $m$ on the incline is $\mu > \tan37^{\circ}=0.75355$. However, the friction coefficient here is $0.75$, which is less. This means that the block $m$ will start to slide down the plane while $M$ falls. This complicates the problem.
It is confusing of the question-setter to choose an angle so close to the critical value, instead of $30^{\circ}$ or $40^{\circ}$.