Question

A mass of 0.5 kg is suspended from a wire, then the length of the wire increases by 3 mm then work done is?

Answer 1

Step 1, Given data

Mass of the suspended body = 0.5 kg

Increases in length $\Delta l=3mm=3\times {10}^{-3}m$

Step 2, Finding the work done

We know that

$Workdone=\frac{1}{2}\times mg\times \Delta l$

Putting all the values

$Workdone=\frac{1}{2}\times \frac{1}{2}\times 9.8\times 3\times {10}^{-3}$

After solving

Work done = $7.35\times {10}^{-3}J$

Hence the total work done is $7.35\times {10}^{-3}J$