The value of potential energy stored in the spring when the block is 6. 20 cm from equilibrium is 1.65 J.
<h3>What is elastic potential energy?</h3>
The elastic potential energy is the energy which is stored in a stretched spring. The elastic potential energy of a spring can be found by the following formula.
![U_s=\dfrac{1}{2}kx^2](https://tex.z-dn.net/?f=U_s%3D%5Cdfrac%7B1%7D%7B2%7Dkx%5E2)
Here, (k) is spring constant and (x) is the displacement of the spring.
The block is pulled to a position xi = 6. 20 cm from equilibrium and released. Thus,
![x=6.20\rm \; cm](https://tex.z-dn.net/?f=x%3D6.20%5Crm%20%5C%3B%20cm)
Divide with 100 to convert its unit in meter,
![x=0.062\rm \; m](https://tex.z-dn.net/?f=x%3D0.062%5Crm%20%5C%3B%20m)
Let the value of spring constant is 860 N/m. Thus, put the values in the formula,
![U_s=\dfrac{1}{2}kx^2\\U_s=\dfrac{1}{2}(860)(0.062)^2\\U_s=1.65\rm\; J](https://tex.z-dn.net/?f=U_s%3D%5Cdfrac%7B1%7D%7B2%7Dkx%5E2%5C%5CU_s%3D%5Cdfrac%7B1%7D%7B2%7D%28860%29%280.062%29%5E2%5C%5CU_s%3D1.65%5Crm%5C%3B%20J)
Thus, the value of potential energy stored in the spring when the block is 6. 20 cm from equilibrium is 1.65 J.
Learn more about the elastic potential energy here;
brainly.com/question/26497164
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