Question

A spring has a natural length of 10 cm. if a 26-n force is required to keep it stretched to a length of 16 cm, how much work w is required to stretch it from 10 cm to 13 cm? (round your answer to two decimal places.)

Answer 1

When a force of F=26 N is applied to the spring, the spring is stretched with respect to its rest position by

`\Delta x=16 cm - 10 cm=6 cm = 0.06 m`

This means that the constant of the spring is

`k= (F)/(\Delta x)= (26 N)/(0.06 m)=433.33 N/m`

The work required to stretch the spring from 10 cm to 13 cm, which corresponds to an elongation of

`\Delta x_2 = 13 cm - 10 cm=3 cm =0.03 m`
is equal to the variation of elastic potential energy of the spring, so:

`W=\Delta U= (1)/(2)k(\Delta x_2)^2= (1)/(2)(433.3 N)(0.03 m)^2=0.19 J`