JEE MAIN - Physics (2003 - No. 64)
A wire suspended vertically from one of its ends is stretched by attaching a weight of $$200N$$ to the lower end. The weight stretches the wire by $$1$$ $$mm.$$ Then the elastic energy stored in the wire is
$$0.2$$ $$J$$
$$10$$ $$J$$
$$20$$ $$J$$
$$0.1$$ $$J$$
Explicação
The elastic potential energy
$$ = {1 \over 2} \times $$ Force $$ \times $$ extension
$$= {1 \over 2} \times 200 \times 0.001 = 0.1\,J$$
$$ = {1 \over 2} \times $$ Force $$ \times $$ extension
$$= {1 \over 2} \times 200 \times 0.001 = 0.1\,J$$