a small block of mass m slides without friction around the loop-the-loop apparatus shown below. (a) if the block starts from rest at a, what is its speed at b? (b) what is the force of the track on the block at b?

The speed of the block at point B is √(6Rg) and the force by the track on the block is 6mg.The mass of the block is m.The point A is at a height of 4R and point B is at a height of R.(a) According to law of conservation of energy, the energy in gravitational field always remains same at every point.Potential energy at A + Kinetic energy at A = Potential energy at B + Kinetic energy at BPotential energy at A is mh4R.Kinetic energy at A is zero.Potential energy at B is mgR.Kinetic energy at A is 1/2mv²Putting all the values,mh4R = 1/2mv² + mgRmg3R = 1/2mv²v = √(6Rg)So, the speed of the block at point B is √(6Rg).(b) The force F of the track on the block is,F = mv²/rv is speed at point B.r is the radius of the track.Putting values,F = 6mg.The force by the track on the block is mg.To know more about law of energy conservation, visit,https://brainly.com/question/166559#SPJ4