Question

Find the work done in stretching a spring of natural length 8 cm, from 10 cm to 13 cm. Assume a force of 6 N is needed to hold it at a length of 11 cm.

Answer 1

$Force=Spring\ constant\ *\ elongation\\ F=k*e\\$

$at\ length=11cm,\\ 6=k*(11-8)\\ k=2Ncm^{-1}\\$

$at\ length=10cm,\\ F_{10cm}=2*(10-8)\\ F_{10cm}=4N\\$

$at\ length=13cm,\\ F_{13cm}=2*(13-8)\\ F_{13cm}=10N\\$

$work\ done\ =\ force\ *\ distance\\ work\ done\ =\ \frac{(4+10)}{2}\ *\ (13-10)*10^{-2}\\ (multiply\ by\ 10^{-2}\ for\ convert\ into\ meters.)\\ \bold {work\ done\ =\ 0.21Nm}$