Determine the energy stored in a spring if it is stretched by 20 cm and the spring constant is 500N/m?

To either increase or decrease the length of a spring from
its equilibrium length work has to be done on the spring. The work done is stored as
potential energy in the spring and when the spring is let free it regains its
equilibrium length.

The force that has to be applied to
change the length of a spring is dependent on the change in length. It is equal to F =
-k*x, where x is the change in length. The negative sign indicates a restoring
force.

Work is force*distance. If the spring has been
extended by x, the work to be done to intend the length further by an infinitesimal
length dx is dW = k*x dx

Here the spring constant is 500
N/m and the spring is extended by 20 cm. The work done is the definite integral
Int[500*x dx], x= 0 to x = 0.20

=> 500*x^2/2, x = 0
to x = 0.2

=>
500*0.2^2/2

=>
250*0.04

=> 10 J

The
energy stored in the spring when it is stretched 20 cm is 10 J