A spring has a spring constant of 1000 N/m and its initial expansion is 10 cm. Calculate the work done in extending the spring from 10 cm to 20 cm.

CONCEPT:

• Elastic potential energy is the energy stored as a result of applying a force to deform an elastic object. 
• The energy is stored until the force is removed and the object regains its original shape, doing work in the process.
• This deformation could involve compressing, stretching, or twisting the object.
• The elastic potential energy is associated with the work done.
• The elastic potential energy is equal to the work done (w) in the spring.

\(⇒ W = {1 \over 2} k x^2\)

Where k is the spring constant and x is the change in the length of spring from the normal position to the compressed position.

• Many objects are specifically designed  to store elastic potential energy inside them,

​For example:

1. The coil spring of a wind-up clock.
2. An archer's stretched bow.
3. A stretched slingshot ready to fire.

EXPLANATION:

Given:

k = 1000 N/m,  x1 = 10 cm, x2 = 20 cm.

• A spring has a spring constant of 1000 N/m and its initial expansion is 10 cm.

Using formula,

\(⇒ W = {1 \over 2} k x^2\)

Where k is spring constant, x = Displacement of the spring.

⇒  Initially at x₁ = 10 cm = 0.1 m.

\(⇒ {W_1} = \frac{1}{2} \times 1000 \times {0.1^2} = 5\;J\)

at x₂ = 20 cm = 0.2 m.

\(⇒ {W_2} = \frac{1}{2} \times 1000 \times {0.2^2} = 20\;J\)

• Work done to extend the spring up to 20 cm.