A block of mass M1 = 0.561 kg sits on top of a block with a mass of M2 = 4.43 kg

Question

A block of mass M1 = 0.561 kg sits on top of a block with a mass of M2 = 4.43 kg. The coefficient of static friction between the two blocks is ? = 0.295. The lower block rests on a frictionless surface and is connected to a spring that has a spring constant of k = 46.1 N/m. The lower block is then displaced by x as shown and undergoes simple harmonic motion. What is the largest amplitude the block can have for the smaller block to remain at rest relative to the larger block?

(Please show your work)

Explanation / Answer

w = ? k / m = ? 46.1 / 4.991 = 3.03 rad/s

a(max) = w^2 . A = 3.03^2 . A

Force needed to give this acceleration to the block M1 = 0.561 . (3.03^2 . A )

Maximum friction force =

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