A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has force constant 815 N/m . The coefficient of kinetic friction between the floor and the block is μk=0.40. The block and spring are released from rest and the block slides along the floor. Required:
What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0100 m.)

v = 0.41 m/s

Explanation:

In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.So, we can write the following general equation, taking the initial and final values of the energies:

       

Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.⇒ Kf = 1/2*m*vf²  (2)The change in the potential energy, can be written as follows:

       

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

Regarding the work done by the force of friction, it can be written as follows:

       

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:Fn = Fg= m*g (5)Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

       

       

Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

   

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