A block of mass m1 = 5.40 kg sits on top of a second block of mass m2 = 16.1 kg,

Question

A block of mass m1 = 5.40 kg sits on top of a second block of mass m2 = 16.1 kg, which in turn is on a horizontal table. The coefficients of friction between the two blocks are s = 0.300 and k = 0.100. The coefficients of friction between the lower block and the rough table are s = 0.500 and k = 0.400. You apply a constant horizontal force to the lower block, just large enough to make this block start sliding out from between the upper block and the table.

determine the acceleration you measure for each block?

Explanation / Answer

Here ,

for the maximum acceleration of the blocks

for the upper block

Using second law of motion

m1 * a = m1 * uk * g

a = 0.10 * 9.8

a = 0.98 m/s^2

the acceleration of the upper block is 0.98 m/s^2

for the lower block

acceleration needed to slide the upper block is

a2 = us * g

a2 = 0.300 * 9.8

a2 = 2.94 m/s^2

the acceleration of the lower block is 2.94 m/s^2

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