Question

spring constant of spring is 5dyne cm^_1and it is elongated from 20cm to22 cm then energy stored in spring is​

Answer 1

Answer:10 ergs

Explanation:

energy stored=1/2 k x^2

                       =1/2 (5)(2)^2

                       =1/2(5)(4)     =5(2)=10 ergs

Answer 2

Answer:

The spring constant of a spring is 5dyne cm^_1and it is elongated from 20cm to22 cm then the energy stored in the spring is​ $10erg$

Explanation:

• The elastic potential energy is the energy stored when a force applied to deform an elastic object

• A spring with elastic constant K compressed or expanded at a distance x from its mean position, then the potential energy stored is

        $PE=\frac{1}{2}Kx^{2}$

From the question, we have

the spring constant $K=5dyne/cm$

extended length of the spring $x=22cm-20cm=2cm$

elastic potential energy

$PE=\frac{1}{2}*5*2^{2} =10erg$