Question

An object is attached to a hanging unstretched ideal and massless spring and slowly lowered to its equilibrium position, a distance of 6.4 cm below the starting point. If instead of having been lowered slowly the object was dropped from rest, how far then would it then stretch the spring at maximum elongation?

Answer 1

Answer:

        h = 12.8 cm

Explanation:

from the equation we are given the following:

distance = 6.4 cm

• when the object is being lowered the weight is equal to the spring force

        weight = spring force

         mg = ky ... equation 1

• potential energy to stretch a spring = work done by the spring

        mgh = 0.5 x k x h^{2} ....equation 2

• substituting equation 1 into equation 2

                kyh =  0.5 x k x h^{2}

                y =  0.5 x h

                2y = h

• where y = 6.4 , the maximum elonf=gation becomes

          h = 2 x 6.4 = 12.8 cm