A 29.0-kg block is resting on a flat horizontal table. On top of this block is r

Question

A 29.0-kg block is resting on a flat horizontal table. On top of this block is resting a 17.0-kg block, to which a horizontal spring is attached, as the drawing illustrates. The spring constant of the spring is 305 N/m. The coefficient of kinetic friction between the lower block and the table is 0.400, and the coefficient of static friction between the two blocks is 0.800. 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, m

Explanation / Answer

1) Spring force can be expressed as = kx

k = spring constant =305N/m, x= distance moved by spring

Frictional force by the 17Kg force = f*m1*g

Since the frctional force comptresses the spring we can sustitute kx = f*m1*g

x= f*m1*g/k = 0.8*17*9.8/305 = 0.436m

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