An ideal massless spring `S` can compressed `1.0` m in equilibrium by a force of `100 N`. This same spring is placed at the bottom of a friction less inclined plane which makes an angle `theta =30^@` with the horizontal. A (10 kg) mass (m) is released from rest at the top of the incline and and is brought to rest momentarily after compressing the spring by `2 m. if `g =10 ms^(-1)`, what is the speed of just before it touches the spring? .

Options

A. (a) `sqrt20 ms^(-1)`
B. (b) `sqrt30 ms^(-1)`
C. (c) `sqrt10 ms^(-1)`
D. (d) `sqrt40 ms^(-1)`

Answer

A

Explanation

`F=kx`
`:. k=(F)/(x) =(100)/(1)`
`=100 N//m`
`E_(i)=E_(f)`
`:. 1/2 xx 10 xx v^(2) =1/2 xx 100 xx (2)^(2)-(10)(10)(2 sin 30^@)`
Solving we get,
`v=sqrt20 m//s`