A 28.0-kg block is resting on a flat horizontal table. On top of this block is r

Question

A 28.0-kg block is resting on a flat horizontal table. On top of this block is resting a 12.0-kg block, to which a horizontal spring is attached, as the drawing illustrates. The spring constant of the spring is 325 N/m. The coefficient of kinetic friction between the lower block and the table is 0.430, and the coefficient of static friction between the two blocks is 0.790. 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. Determine the amount by which the spring is compressed at the point where the upper block begins to slip on the lower block.

Explanation / Answer

since the blocks are movinf at constant speed net force = 0

so,

so, upper block will start slipping after spring forces surpasses the frictional force

kx = 0.790 * 12 * 9.8

325 * x = 0.790 * 12 * 9.8

x = 0.286 m

the amount by which the spring is compressed at the point where the upper block begins to slip on the lower block = 0.286 m

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