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The tension in a horizontal spring is directly proportional to the extension (1 mark)
of the spring. The energy stored in the spring at extension X is E. What is
the work done by the spring when its extension changes from X to X/4?

Sagot :

Hamzaahmedsidn

This question involves the concepts of work done and elastic potential energy.

The work done by the spring when its extension changes from X to X/4 is "0.56 E".

The work done by the spring is equal to the elastic potential energy stored in the spring, which is given as follows:

[tex]W=E\\W=E=\frac{1}{2}kX^2---- eqn(1)[/tex]

where,

W = work done = ?

E = elastic potential energy

k = spring constant

X = extension

Now, the spring moves to an extension of X/4, so the change in extension will be:

[tex]\Delta X = X-\frac{X}{4}\\\\\Delta X = \frac{3X}{4}[/tex]

Hence, the work done will become:

[tex]W=\frac{1}{2}K\Delta X^2\\\\W=\frac{1}{2}K(\frac{3X}{4})^2\\\\W=\frac{1}{2}KX^2(0.56)\\\\using\ eqn(1):[/tex]

W = 0.56 E

Learn more about Elastic Potential Energy here:

https://brainly.com/question/156316?referrer=searchResults

Attached picture shows Elastic Potential Energy.

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