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

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Best answer

Your answer is twice the given answer. Unless the given answer is wrong, you have not found the minimum mass.

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.

Why maximum tension is 2Mg

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 $40^{\circ}$.

It is confusing of the question-setter to choose an angle so close to the critical value, instead of $30^{\circ}$ or $40^{\circ}$.

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