Ablock of mass 200g is sitting on top of another block of three times that mass which is on a horizontal frictionless surface and is attached to a horizontal spring. the coefficient of static friction between the blocks is 0.2. the lower block is pulled until the attached spring is stretched by 5.0cm and is then released from rest. find the maximum value of the spring constant for which the upper block does not slip on the lower block.

k = 39.2 N / m

Explanation:

The 200 g block is accelerated by the force of friction between the blocks. Let's use Newton's second law

    N- W = 0

    N = W

    fr = ma

   μ N = ma

   μ mg = ma

   a=μ g

Let's look for the acceleration of the largest block that has oscillatory movement

    x = A cos (w t)

   A = 0.05 m

The maximum acceleration is  cos wt = ±1

    a = A w2

    a = A k / m

We substitute and calculate

  μ g = A k / M

   k = μ g M / A

The mass that performs the oscillation is the mass of the two bodies

   M = m1 + m2

   k = 0.2 9.8 (0.800+ 0.200) /0.05

   k = 39.2 N / m

answer; ///(74) degrees is your answer..