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`.

Explanation

Applying work energy theorem for motion from (a) to (b)
`W_("gravity")+W_("friction")+W_("spring")=DeltaK.E=(1)/(2)m(0-0)=0`
`therefore 20xx5sin37^(@)-mu(20cos37^(@))5-(1)/(2)k[(0.2)^(2)-0]=0" "....(i)`
Applying work energy for motion form (b) to (c)
`-20xx1xxsin37^(@)-mu(20cos37^(@))xx1-(1)/(2)k[0-(0,2)^(2)]=0" ".....(ii)`
Adding equation (i) and (ii)
`therefore-20(5-1)xx(3)/(5)-mu(20xx(4)/(5))(5+1)=0`
`impliesmu=0.5`
Putting this value in equation (i), we get
`k = 1000 N//m`
Here velocity is maximum at equilibrium since before this, spring force was less than the weight of the block and the block was accelerating and after this , the spring force is greater than the weight thus retarding the block to zero velocity upto the lowest position.