Question

A spring of force constant $800N/m$ has an extension of $5cm$. The work done in extending it from $5cm$ to $15cm$ is

• A. 16 J
• B. 8 J
• C. 32 J
• D. 24 J

Answer 1

Answer: (B)

The work done is stored as elastic potential energy in the spring.
It is given by
$W=\frac{1}{2}\times k\left(x_2{}^2-x_1{}^2\right)$
$x_1=5cm$
$x_2=15cm$
$k=800\frac{N}{m}=\frac{800}{100}\frac{N}{cm}=8\frac{N}{cm}$
$W=\frac{1}{2}\times 8\left(15^2-5^2\right)=\frac{1}{2}\times 8(225-25)$
$=\frac{1}{2}\times 8\times 200=800\frac{N}{cm}$
$W=800\frac{N}{cm}=8\frac{N}{m}=8J$