A small body slides down a smooth uneven surface from a height H , which eventually emerges into a circular loop of radius R(

Step-by-step solution

Given, height through which body slides =H Radius of circular loop =R Force on body A, F_A= 2 weight (w) Let, velocity of body at A=v_A Acceleration due to gravity =g m v_A^2 R = 2 w= 2 m g v_A^2= 2 Rg By using law of conservation of energy Energy at position (P)= Energy at position (A) array rlrl & & m g H & = 1 2 m v_A^2+m g R & 2 g H & =v_A^2+2 g R= 2 g R+2 g R=( 2 +2) g R & & H & = ( 1 2 +1 ) R=1.7 R=1.5 R= 3 2 R array