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3 years ago

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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?

1 answer:

myrzilka [38]3 years ago

3 0

<h2>Answer:</h2>

4E

<h2>Explanation:</h2>

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 = $\frac{1}{2}kx^2$ --------------------(i)

Where;

U = potential energy stored

k = spring constant of the material

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

<em>From the first statement;</em>

<em>when elongation (x) is 4cm, energy stored (U) is E</em>

<em>Substitute these values into equation (i) as follows;</em>

E = $\frac{1}{2}k(4)^2$

E = 8k

<em>Make k subject of the formula</em>

k = $\frac{E}{8}$ [measured in J/cm]

<em>From the second statement;</em>

<em>It is stretched by 4cm.</em>

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 = $\frac{E}{8}$ into equation (i) as follows;

U = $\frac{1}{2}\frac{E}{8} (8)^2$

U = $\frac{1}{2}{8E}$

U = ${4E}$

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

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