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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.

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whitneytr12

Answer:

[tex]W=5.16 J[/tex]  

Explanation:

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

[tex]F=-k\Delta x[/tex]

[tex]30.8=-k*0.177[/tex]

[tex]k=|-\frac{30.8}{0.177}|[/tex]

[tex]k=174 N/m[/tex]

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

[tex]W=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})[/tex]

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

x1 is the initial distance (x1 = 0.177 m)

[tex]W=\frac{1}{2}*174(0.301^{2}-0.177^{2})[/tex]

[tex]W=5.16 J[/tex]    

I hope it helps you!

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