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In a wire, when elongation is 4 cm energy stored is E. if it is stretched by 4 cm, then what amount of elastic potential energy will be stored in it?

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2 votes

Answer:

4E

Step-by-step explanation:

The elastic potential energy of an elastic material (e.g a spring, a wire), is the energy stored when the material is stretched or compressed. It is given by

U =
(1)/(2)kx^2 --------------------(i)

Where;

U = potential energy stored

k = spring constant of the material

x = elongation (extension or compression of the material).

From the first statement;

when elongation (x) is 4cm, energy stored (U) is E

Substitute these values into equation (i) as follows;

E =
(1)/(2)k(4)^2

E = 8k

Make k subject of the formula

k =
(E)/(8) [measured in J/cm]

From the second statement;

It is stretched by 4cm.

This means that total elongation will be 4cm + 4cm = 8cm.

The potential energy stored will be found by substituting the value of x = 8cm and k =
(E)/(8) into equation (i) as follows;

U =
(1)/(2)(E)/(8) (8)^2

U =
(1)/(2){8E}

U =
{4E}

Therefore, the potential energy stored will now be 4 times the original one.

User Jinlong
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