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

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.