Figure shows a spring fixed at the bottom end of an incline of inclination `37^0`. A small block of mass 2 kg starts slipping down the incline from a point 4.8 m away from the spring. The block compresses the spring by 20 cm, stops momentarily and then rebounds through a distance of 1 m up the incline. Find a. the frictioin coefficient between the plane and the block and b. the spring constant of the spring. Take `g=10 m/s^2`.

Answer

A, B

Explanation

`m=2 kg, S_1=4.8m,x=20 cm=0.2m`
`S_2=1m`
`sin37^0=0.60=3/5`
`theta=37^0`
cos 37^0=0.80=4/5=10m/sec^2`

Applying work energy principle for downward motion of the body
`0-0=mgsin37^0(x+4.8)-muRx5-1/2kx62`
`20xx(0.80)xx5-1/2k(0.2)^2=0`
`rarr 60-80mu-0.02k=0`
rarr `80mu+0.02k=60`
Similarly for the upward motion of the body the equation is
`0-0=(-mgsin37^0)-muRx1+1/(2k)(-2)^2`
`rarr 20xx(0.06)xx1-muxx20xx(0.80)xx1-1/2k(0.2)^2`
`rarr 12-16mu+0.02k=0`
`Adding equation i and equation ii we get
`96mu=48`
`rarr mu=0.5`
Now putting the value of `mu` in equation i,
`k=1000 N/m`