An ideal massless spring S can compressed `1.0` m in equilibrium by a force of `1000 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 the rest at top of the inclined plane and is brought to rest momentarily after compressing the spring by `2.0 m.` the distance through which the mass moved before coming to rest is.

Options

A. `8 m`
B. `6 m`
C. `4 m`
D. `5 m`

Answer

C

Explanation

From `F =kx`

`k = (F)/(x)=(100)/(1.0)`
Decrease in gravitational potential energy `= increase in spring potential energy
`rArr 10 xx 10 xx (d + 2) sin 30^(@) =1/2 xx 100 xx (2)^(2)`
Solving, we distance covered before coming momentarily to rest
`= d + 2 = 4m`.