The system is released from rest with spring intially in its natural length. If mass of the block m = 10 kg. and spring constant `k = 100 N//m`, then maximum extension in spring is :

Options

A. 1 m
B. `(1)/(2) m`
C. 2 m
D. 2.5 m

Answer

A

Explanation

Let maximum elongation in spring be X, then block descends through `(x)/(2)`

BY WET
`"mg"(x)/(2)-(1)/(2)kx^(2)=0rArr X=(mg)/(k)=(10xx10)/(100)=1m`