Question

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?

Answer 1

Answer:

4E

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

Where;

U = potential energy stored

k = spring constant of the material

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

From the first statement;

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

Substitute these values into equation (i) as follows;

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

E = 8k

Make k subject of the formula    

k = $\frac{E}{8}$   [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 = $\frac{E}{8}$ 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.