When 1 N force is applied increase in length of the spring is 1 cm . Find elastic potential energy stored during this in it .

Options

A. `10 xx10^(-3)J `
B. `10^(-3)` J
C. `5xx10^(-3) J `
D. `20 xx10^(-3) J `

Answer

C

Explanation

From ` F = - kx , k = F/x 1/(10^(-2)) = 10^(2) `N/m
(neglecting negative sign )
` U = 1/2 kx^(2) = 1/2 xx10^(2) xx (10^(-2))^(2)`
` = 5xx10^(-3) J ` .