Question

a mass of 0. 5 kg is suspended from wire ,then length of wire increase by 3mm find out work done:

Answer 1

The work done is 7.5x$10^{-3}$ j

Explanation:

Consider a wire of length L, area of cross section A. A load is hung which creates a force f and elongates the wire by l

then , y = stress/strain = (f/A)/(x/L) = fL/Ax

f = $\frac{yAx}{L}$..........(1)

Work done by external force in elongating the wire

dW = fdx...............(2)

dW = $\frac{yAx}{L}$dx

Taking integration

∫dW = $\frac{yA}{L}\int\limits^l_0 {x} \, dx$

W = $\frac{yA}{L}[\frac{x^{2} }{2}]$ limit 0 - l

W = $\frac{yA}{L}[\frac{l^{2} }{2}-0]$.....................(3)

Now from equation (1) we have f = $\frac{yAx}{L}$

When wire is fully stretched x = 1

f = Mg

Mg = yAl/L.........(4)

from equation 3 and 4th

W = (mg)l/2

W =(1/2)Fl.

Using the formula from proof.

W = 1/2Fl

W = $\frac{1}{2}x(0.5x10)(3x10^{-3})$

W = $\frac{15}{2}x10^{-3}$

Then the workdone will be W = 7.5x$10^{-3}$J