What Amplitude Causes Slippage in a Two-Block Spring System?

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SUMMARY

The discussion focuses on determining the amplitude of simple harmonic motion in a two-block spring system that causes the smaller block (mass m=1.0 kg) to slip over the larger block (mass M=10 kg). The spring constant is k=200 N/m, and the coefficient of static friction between the blocks is μs=0.40. The derived equations indicate that the acceleration of the smaller block is a=-18.2Acos(4.26t), where A is the amplitude. The critical point is that only the static friction coefficient is necessary to find the amplitude at which slipping occurs, rendering the lengths of the blocks irrelevant.

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Two blocks (m=1.0 kg and M=10 kg) and a string (k=200 N/m) are arranged on a horizontal, frictionless surface (in the picture m rests atop M). The coefficient of static friction between the two blocks is .40. What amplitude of simple harmonic motion of the spring-blocks system puts the smaller block on the verge of slipping over the larger block?

acom=F/(mA+mB)

F=-kAcos(ωt) and ω=√(k/(mA+mB)= 4.26

so F=-200Acos(4.26t)

acom=-200Acos(4.26t)/11
=-18.2Acos(4.26t)

This is all I got so far. What I am confused about is that μk is not given. So isn't it impossible to figure out the acceleration of the top block once static friction is overcome? I also am not given the lengths of the blocks.
 
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The question means, at what amplitude would it start to slip? For that you only need static friction coefficient, and the block widths are irrelevant.
 

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