JEE MAIN - Physics (2004 - No. 53)
A wire fixed at the upper end stretches by length $$l$$ by applying a force $$F.$$ The work done in stretching is
$$2Fl$$
$$Fl$$
$${F \over {2l}}$$
$${{Fl} \over 2}$$
Erläuterung
Work done by constant force in displacing the object by a distance $$\ell $$.
= Potential energy stored
$$ = {1 \over 2} \times $$ Stress $$ \times $$ Strain $$ \times $$ Volume
$$ = {1 \over 2} \times {F \over A} \times {l \over L} \times AL$$
$$ = {1 \over 2}Fl$$
= Potential energy stored
$$ = {1 \over 2} \times $$ Stress $$ \times $$ Strain $$ \times $$ Volume
$$ = {1 \over 2} \times {F \over A} \times {l \over L} \times AL$$
$$ = {1 \over 2}Fl$$