Question

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 30.8 N, the spring is stretched by 17.7 cm from its equilibrium length. Calculate the additional work required by F to stretch the spring by an additional 12.4 cm from that position.

Answer 1

`W=5.16 J`

Step-by-step explanation:

Using the Hooke's law we can find the elasticity constant:

`F=-k\Delta x`

`30.8=-k*0.177`

`k=|-(30.8)/(0.177)|`

`k=174 N/m`

Now, we know that the work done is equal to the elastic energy, so we will have:

`W=(1)/(2)k(x_(2)^(2)-x_(1)^(2))`

x2 is the final distance (x2 = 0.177+0.124 = 0.301 m)

x1 is the initial distance (x1 = 0.177 m)

`W=(1)/(2)*174(0.301^(2)-0.177^(2))`

`W=5.16 J`

I hope it helps you!