Question

a mass of 0.5 kg is suspended from wire,then length of wire increased by 3mm, then work done





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Answer 1

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Answer 2

The work done is  $7.5\times10^-^3J$.

Explanation:

The work done on the wire during stretching is given as

$W=\frac{1}{2}FL$

Here, F is the force and L is the extension in the wire.

We can also write,

$W=\frac{1}{2}mg\times L$

Given m= 0.5 kg, L = 3 mm and  take g = 9.8 m/s^2

Substitute the given values in above formula, we get

$W=\frac{1}{2}(0.5kg\times9.8m/s^2\times3\times10^-^3m)$

$W=7.5\times10^-^3J$

Thus, work done is  $7.5\times10^-^3J$

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