A 30.0-kg block is resting on a flat horizontal table. On topof this block is re

Question

A 30.0-kg block is resting on a flat horizontal table. On topof this block is resting a 15.0-kg block, to which a horizontalspring is attached, as the drawing illustrates. The spring constantof the spring is 335 N/m. Thecoefficient of kinetic friction between the lower block and thetable is 0.620, and the coefficient ofstatic friction between the two blocks is 0.895. A horizontal force F is applied to thelower block as shown. This force is increasing in such a way as tokeep the blocks moving at a constant speed. At the point where theupper block begins to slip on the lower block determine thefollowing. (a) the amount by which the spring is compressed
1 m

(b) the magnitude of the force F
2 N PLEASE PROVIDE STEPS! THANKS IN ADVANCE

Explanation / Answer

   let us take lower block as 1 and upper blockas 2, then
   m1 = 30.0 kg
   m2 = 15.0 kg
   k = 335 N / m
   k = 0.620
   s = 0.895
   as shown in the figure when the force is applied tothe lower block the kinetic frictional force
   Fk1 on the bottom block is givenby
   Fk1 = k (m1 +m2) g
   as a result static frictional force is applied on thetop block given by
   Fs2 = s m2g
   when we apply newtons second law to the bottom weget
   F - Fk1 - k x = 0
   and for the top block we get
   Fs2 - k x = 0
(a)
   so the amount by which the spring is compressed willbe
   x = Fs2 / k
      = sm2 g / k
      = ............. m
(b)
   the magnitude of the force F willbe
   F = k x + Fk1
      = k x + k(m1 + m2) g
      = ................N

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