An ideal massless spring S can be compressed 1 m by a force of 100 N in equilibrium. The same spring is placed at the bottom of a frictionless plane inclined at `30^(@)` to the horizontal. A 10 kg block M is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring by 2 m. If `g = 10 m//s^(2)`, the speed of mass just before it touches the spring is `sqrt(10x) m//s`. Find value of x?

Options

A. `sqrt20ms^(-1)`
B. `sqrt30ms^(-1)`
C. `sqrt10ms^(-1)`
D. `sqrt40ms^(-1)`

Answer

A

Explanation

(a) F=kx
`[(v^(3))/(3)]_(v)^(2v)=(Ps)/(m)" or "s=(7mv^(3))/(3P)`
Now from energy conservation, between natural length of spring anf its maximum compression state.
`(1)/(2)mv^(2)+mgh=(1)/(2)kx_(max)^(2)`
`v=sqrt((kx_(max)^(2))/(m)-2gh)=sqrt(((100)(2)^(2))/(10)-(2)(10)(1))`
`=sqrt(20) ms^(-1)`