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Q.
A spring of force constant $800\, N / m$ has an extension of $5\, cm$. The work done in extending it from $5\, cm$ to $15\, cm$ is
AIEEEAIEEE 2002Work, Energy and Power
Solution:
The work done is stored as elastic potential energy in the spring.
It is given by
$ W =\frac{1}{2} \times k \left( x _{2}{ }^{2}- x _{1}{ }^{2}\right) $
$x _{1}=5 cm$
$x _{2}=15 cm$
$k =800 \frac{ N }{ m }=\frac{800}{100} \frac{ N }{ cm }=8 \frac{ N }{ cm } $
$W =\frac{1}{2} \times 8\left(15^{2}-5^{2}\right)=\frac{1}{2} \times 8(225-25) $
$=\frac{1}{2} \times 8 \times 200=800 \frac{ N }{ cm } $
$W =800 \frac{ N }{ cm }=8 \frac{ N }{ m }=8 J $