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The right end of a relaxed standard spring is at the origin; the left end is clamped at some point on the negative x-axis. Holding the spring’s right end at location x=5 cm requires a force of N. Find the work (in Joules) required to stretch the spring from 5 cm to 8 cm.

The Right End Of A Relaxed Standard Spring Is At The Origin The Left End Is Clamped At Some Point On The Negative Xaxis Holding The Springs Right End At Locatio class=

Answer :

Lhannearaujogo

The work required to stretch the spring is 350J.

Hooke's Law

This law can be written by the formula: F=kx, where:

F= elastic force (N)

k= spring constant (N/m)

x= linear displacement (m)

Spring Work

For finding the spring work in J, you should apply the formula  [tex]W=\frac{k*(\Delta x)^2}{2}[/tex], where:

W= work (J)

k= spring constant (N/m)

x= difference between the linear displacements (m)

The question gives:

x=5 cm=0.05m requires a force of 1.4N

x=8 cm =0.08m

  • Step 1 - First, you should find the spring constant from Hooke's law, for x=5 cm and F=1.4N.

                                          [tex]F=kx\\ 1.4=k*0.05\\ \\ k=\frac{1.4}{0.05} =28\frac{N}{m}[/tex]

  • Step 2 - Now you can apply the formula for spring work.

                                       [tex]W=\frac{k*(\Delta x)^2}{2}\\ \\ W=\frac{28*(8-3)^2}{2} \\ \\ W=\frac{28*(5)^2}{2}\\ \\ W=\frac{28*25}{2}\\ \\ W=\frac{700}{2}=350J[/tex]

Read more about the spring work here:

https://brainly.com/question/3317535

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