Ablock (0.50 kg) is attached to an ideal spring with a spring constant of 80 n/m, oscillating horizontally on a frictionless surface. the total mechanical energy is 0.12 j. (a) what is the greatest extension of the spring from its equilibrium length? (b) now the block is traveling 2.00 m/s, and brought to rest by compressing a very long spring of spring constant 800.0 n/m. how much does the spring compress?

a) x =  5.48 10⁻² m and b)  0.05 m

Explanation:

a) For a system in oscillatory motion the mechanical energy conserves and is described by the equation

     Em = ½ k A²

Where k is the spring constant and at the amplitude of the movement

When the spring has the greatest extent, the kinetic energy is zero

     Em = U = ½ k x²

Therefore, the amplitude of the movement is the same amplitude of the spring

Let's calculate

    A = √ (2Em / k)

    A = √ (2 0.12 / 80)

   A = 0.0548 m = 5.48 10⁻² m

b) In this case the spring has kinetic energy that becomes elastic potential energy, let's calculate the mechanical energy before and after compressing the spring

Initial

      Em = K = ½ m v²

Final

     Em = Ke = ½ k x²

     ½ m v² = ½ k x²

     x = √(m/k) v

     x = 2 √(0.50 /800.0)

     x = 0.05 m