A long spring is stretched by 2 cm and its potential energy is U. If the spring is stretched by 10 cm, then its potential energy would be:

Concept:

Elastic potential energy: It is the potential energy created when an object undergoes an elastic deformation due to a force applied to it.

• This energy is stored in the object as long as the force is in action and the object goes back to its original shape when the force is removed.
• A spring undergoes a similar situation when a force acting on it causes a displacement due to elastic deformation.

The potential energy (U) required to stretch a string by x distance is given by

\(\Rightarrow U = \frac{1}{2}kx^2\)

where k is the spring constant.

Calculation:

Given:

Let the potential energies of the spring be U₁ and U₂ when they are stretched be x₁ and x₂ respectively.

Given that: U₁ = U

When x₁ = 2 cm:

\(\Rightarrow U_1 = \frac{1}{2}kx_1^2 = \frac{1}{2}k(2)^2\)    ------ (1)

When x₂ = 10 cm:

\(\Rightarrow U_2 = \frac{1}{2}kx_2^2 = \frac{1}{2}k(10)^2\)    ------- (2)

On dividing equations (1) and (2), we get

\(\Rightarrow \frac{U_1}{U_2} = \frac{\frac{1}{2}k(2)^2}{\frac{1}{2}k(10)^2} = \frac{2^2}{10^2} = \frac{1}{25}\;\)

⇒ U₂ = 25U₁ = 25U