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? .

Explanation

(a) Let total distance moved by the block is
`S=(l+2)m`,
where `l` is the distance moved by the block before touching the spring.
Now, work done by gravity on the block is
`W_(g)=mg S " " sin theta = 10 xx 10 xx S sin 30J`
`rArr W_(g)=50 S J ` ...(1)
Work done by spring on the block is , `W_(s)=1/2 kx^(2)`
Here, `K=100N//m` and `x=2m`
`rArr W_(s)=1/2xx 100 xx 4J rArr W_(s)=200 J` ...(2)
Total work done `W=W_(g)+W_(s)`
`rArr W=(50 S-200)J`
Since change in K.E. of the block is zero.
as `W=DeltaK.E.`
`rArr 50 S-200=0`
`rArr S=4m`

(b) As `S=l+2 rArr l=S-2=2m`
Work done by gravity over this path length is
`W_(g)=mgxx2 sin theta = 10 xx 10 xx 2 xx 1/2=100J " " [ :.W_(g)=DeltaK.E.]`
`rArr 100 = 1/2 mv^(2)-0 rArr v^(2)=(100xx2)/(10)=20`
`rArr v=2sqrt(5)m//s`