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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4E

Explanation:

From the question given above, the following data were obtained:

Initial elongation (e₁) = 4 cm = 4/100 = 0.04 m

Initial energy (E₁) = E

Final elongation (e₂) = 0.04 + 0.04 = 0.08 m

Final energy (E₂) =?

The energy stored in a s spring is given by:

E = ½Ke²

Where

E => is the energy

K => is the spring constant

e => is the elongation

From:

E = ½Ke²

Energy is directly proportional to the elongation. Thus,

E₁/e₁² = E₂/e₂²

With the above formula, we can obtain the final energy as follow:

Initial elongation (e₁) = 0.04 m

Initial energy (E₁) = E

Final elongation (e₂) = 0.08 m

Final energy (E₂) =?

E₁/e₁² = E₂/e₂²

E / 0.04² = E₂ / 0.08²

E / 0.0016 = E₂ / 0.0064

Cross multiply

0.0016 × E₂ = 0.0064E

Divide both side by 0.0016

E₂ = 0.0064E / 0.0016

E₂ = 4E

Therefore, the final energy is 4 times the initial energy i.e 4E

d.7.5cm is your answer