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

Here the work done in stretching a spring is stored as the potential energy of the spring., i.e., `(1//2) kx^(2)`.
Thus
`W = (1)/(2) kx_(2)^(2) -(1)/(2) kx_(1)^(2) =(1)/(2) k(x_(2)^(2) - x_(1)^(2))`
=`(1)/(2) xx 800[((15)/(100))^(2) -((5)/(100))^(2)] = 8J`.