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

Answer:

$W=5.16 J$  

Explanation:

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

$F=-k\Delta x$

$30.8=-k*0.177$

$k=|-\frac{30.8}{0.177}|$

$k=174 N/m$

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

$W=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})$

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

x1 is the initial distance (x1 = 0.177 m)

$W=\frac{1}{2}*174(0.301^{2}-0.177^{2})$

$W=5.16 J$    

I hope it helps you!