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

Options

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

Answer

B

Explanation

(b) `W_(1)=(1)/(2)k(5)^(2)xx10^(-4) implies W_(2)=(1)/(2)k(15)^(2)xx10^(-4)`
where `W_(1)` and `W_(2)` are the done in extending spring to 5 cm and 15 cm respectively.
`W_(1to2)=W_(2)-W_(1)`
`=(1)/(2)k[(15)^(2)-(5)^(2)]xx10^(-4)=(1)/(2)xx800(200)xx10^(-4)=8 J`