Question

The tension in a horizontal spring is directly proportional to the extension (1 mark)

Answer 1

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:

$W=E\\W=E=\frac{1}{2}kX^2---- eqn(1)$

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:

$\Delta X = X-\frac{X}{4}\\\\\Delta X = \frac{3X}{4}$

Hence, the work done will become:

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

W = 0.56 E

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