9. A 31.0-kg block is resting on a flat horizontal table. On top of this block i

Question

9. A 31.0-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, as the drawing illustrates. The spring constant of the spring is 335 N/m. The coefficient of kinetic friction between the lower block and the table is 0.410, and the coefficient of static friction between the two blocks is 0.740. 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 (a) Determine the amount by which the spring is compressed at the point where the upper block begins to slip on the lower block. lWW m

Explanation / Answer

m1 = 31 kg

m2 = 15 kg

k = 335 N/m

u = 0.410

f = 0.740

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

k*x = f*m2*g ---> x = f*m2*g/k = 0.740*15*9.8/335 = 0.325 m

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