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, as the drawing illustrates. The spring constant of the spring is 330 N/m. The coefficient of kinetic friction between the lower block and the table is 0.640, and the coefficient of static friction between the two blocks is 0.905. A horizontal force vector F 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 the following

Explanation / Answer

Constant speed ---> a = 0 . Then:
a).k x = f m1 g ---> x = f m1 g/k = 0.905*15*9.8/330= 0.4031 m
b) F - k x - f" (m1+m2)g = 0
F = [f m1 + f"(m1+m2)]g = [0.905*15 + 0.64*45]*9.8 = 415.275 N

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