A small body of mass m = 0.5 kg is allowed to slide on an inclined frictionless track from rest position as shown in the figure. (g = 10 m/s 2 )
If h is double of that minimum height required to complete the loop successfully, calculate resultant force
on the block at position H in newton
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(F net = 30)
Sol. (i) for complete the loop minimum velocity at lowest point is v = 
from energy conservation
mv 2 = mgh
m ( 
) 2 = mgh 
h = 
r Ans.
(ii) h is double then velocity at h position is -
mg2h – mg 2r = 
mv 2 (from energy conservation)
v = 
Normal reaction at highest point.
F R = N + mg = 
F R = 6 mg Ans.
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