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

Answer 2

When a spring is stretched or compressed, work is done. In the spring, elastic potential energy is stored. The spring should be stretched up to 12.8 cm.

What is the potential energy of spring?

When a spring is stretched or compressed, work is done. In the spring, elastic potential energy is stored.

The work done is equal to the elastic potential energy stored, assuming no inelastic deformation has occurred.

According to the law of conservation of energy weight get converted into the spring force;

Weight = Spring force

$\rm mg = ky$

The potential energy of the spring is equal to the work done on the spring;

Potential energy = Work done

$\rm mgh = 0.5 \times k \times h^{2} \\\\ kyh = 0.5 \times k \times h^{2} \\\\ y = 0.5 \times h\\\\ 2y = h$

The given value of y is 6.4 cm. Then,

$\rm h=2y \\\\ \rm h=2 \times 6.4 \\\\ \rm h= 12.8\ cm$

Hence the spring should be stretched up to 12.8 cm.

To learn more about the potential energy of the spring refer to;

https://brainly.com/question/2730954