Question

The block is pulled to a position xi = 6. 20 cm from equilibrium and released. Find the potential energy stored in the spring when the block is 6. 20 cm from equilibrium

Answer 1

The value of potential energy stored in the spring when the block is 6. 20 cm from equilibrium is 1.65 J.

What is elastic potential energy?

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.

$U_s=\dfrac{1}{2}kx^2$

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,

$x=6.20\rm \; cm$

Divide with 100 to convert its unit in meter,

$x=0.062\rm \; m$

Let the value of spring constant is 860 N/m. Thus, put the values in the formula,

$U_s=\dfrac{1}{2}kx^2\\U_s=\dfrac{1}{2}(860)(0.062)^2\\U_s=1.65\rm\; J$

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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