A 30.0-kg block is resting on a flat horizontal table. On top of this block is r

Question

A 30.0-kg block is resting on a flat horizontal table. On top of this block is resting a 15.0-kg block, to which a horizontal spring is attached. The spring constant of the spring is 285 N/m. The coefficient of kinetic friction between the lower block and the table is 0.546, and the coefficient of static friction between the two blocks is 0.829. A horizontal force is applied to the lower block as shown. This force is increasing in such a way as to keep the blocks moving at a constant speed. At the point where the upper block begins to slip on the lower block, determine (a) the amount by which the spring is compressed and (b) the magnitude of the force

Explanation / Answer

mass of the block= 30.0-kg

mass of resting block=15.0-kg

The spring constant of the spring= 285 N/m

The coefficient of kinetic friction between the lower block and the table = 0.546

coefficient of static friction between the two blocks = 0.829.

Constant speed ---> a = 0 . Then:

We know that,k x = f m1g

=>x = f m1 g/k = 0.829*15*9.8/285= 0.427 m

b) F - k x - f ' (m1+m2)g = 0

F = [f m1 + f'(m1+m2)]g = [0.829*15 + 0.546*45]*9.8 = 362.649 N

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