Question

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

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

Answer 1

The correct option is B. 8 J.

Given,

Initial extension = $x_1=5cm$

Final extension = $x_2=15cm$

Spring constant = $k=800N/m=\left(800/100\right)N/cm=8N/cm$

The work done is stored as elastic potential energy in the spring.

It is given by

$W=\frac{1}{2}k(x_2^2-x_1^2)$

$\Rightarrow W=\frac{1}{2}k(x_2^2-x_1^2)=\frac{1}{2}\times 8\times ({15}^2-5^2)=800N/cm=\left(800/100\right)N/m=8J$.