Question

When the load applied to a suspended wire is increased from 3 kg-wt to 5 kg-wt; the elongation increases from 0.6 mm to 1 mm. How much work is done during the extention of the wire.

Answer 1

Initial load, $P_1=3kg.wt=30N$

$\Rightarrow$ Final load, $P_2=5kg.wt=50N$

$\Rightarrow$ Initial deformation, $S_1=0.6mm$

$\Rightarrow$ Final deformation, $S_2=1mm$

$\Rightarrow$ Initial W.D. , $W_1=\frac{1}{2}.P_1{S}_1$

$=\frac{1}{2}\times 30\times 0.6$

$=9N.mm$

Hence, the answer is $9N.mm.$