A block M1 of mass 16.5 kg sits on top of a larger block M2 of mass 26.5 kg whic

Question

A block M1 of mass 16.5 kg sits on top of a larger block M2 of mass 26.5 kg which sits on a flat surface. The kinetic friction coefficient between the upper and lower block is 0.400. The kinetic friction coefficient between the lower block and the flat surface is 0.100. A horizontal force = 92 N pushes against the upper block, causing it to slide. The friction force between the blocks then causes the lower block to slide also. Find the magnitude of the acceleration of the upper block. Find the magnitude of the acceleration of the lower block.

Explanation / Answer

Let acceleration of upper block = A and that of lower block be a 15.5*A = 94 - 0.435*15.5*9.8 or A A = (94/15.5) - 4.263 = 1.80 m/s^2 Also a = [0.435*15.5*9.8 - 0.135*25.5*9.8]/25.5 = 1.268 or 1.27 m/s^2 So acceleration of upper block is 1.80 m/s^2 and that of lower block = 1.27 m/s^2

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