The force constant of a weightless spring is `16 N m^(-1)`. A body of mass `1.0 kg` suspended from it is pulled down through `5 cm` and then released. The maximum energy of the system (spring + body) will be

Options

A. `2 xx 10^(-2)`J
B. `4 xx 10^(-2)`J
C. `8 xx 10^(-2)`J
D. `16 xx 10^(-2)`J

Answer

A

Explanation

`omega = sqrt((K)/(m)) = sqrt((16)/(1)) = 4 rads^(-1)`
Now, `K_(max) = (1)/(2) m omega^(2)A^(2)`
`= (1)/(2) xx 1 xx (4)^(2) xx (5 xx 10^(-2))^(2)`
`= 2 xx 10^(-2)J`