Question

A block of mass 0.2kg is attached to a hanging spring whose spring constant is 20N/m. The block is lifted so that the spring is at its unstretched state (The block is 1.2m above the ground). Then the block is released. (a) Find the initial potential energy of the block. (b) How far will the spring stretch when the block is released?

Answer 1

Answer:

a) Em₀ = 2.35 J , b)  y = 0.98 m

Explanation:

Let's use that the mechanical energy of the system remains constant, for this we look for the energy at two points the highest and the lowest.

Higher

     Em₀ = U = mg y₀

Lower

     $E_{mf}$ = U + Ke = m g $y_{f}$ + ½ k y²

a)   Em₀ = m g y₀

     Em₀ = 0.2 9.8 1.2

     Em₀ = 2.35 J

Regarding the ground, where the potential energy is zero

b)   Em₀ = $E_{mf}$

    mg y₀ = mg  $y_{f}$ + ½ k y²

In these equations the three heights are related

    y = y₀ -  $y_{f}$

   mg y₀ - mg  $y_{f}$ = ½ k y²

   mg y = ½ ky²

   y = 2mg / k

   y = 2 0.2 9.8 / 20

  y = 0.98 m