A small block with mass 0.0475 kg slides in a vertical circle of radius 0.0730 m...
A small block with mass 0.0475 kg slides in a vertical circle of radius 0.0730 m on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block's path, the normal force the track exerts on the block has magnitude 3.70 N. What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?
Homework Answers
Velocity, v = sqrt(0.073 x (3.7 - (0.0475 x 9.8))/0.0475) = 2.23 m/s
Using law of conservation of energy,
Vf = sqrt[2.23^2 - 2 x 9.8 x (2 x 0.073)] = 1.45 m/s
Force, N = (0.0475 x 1.45^2/0.073) - (0.0475 x 9.8) = 0.90 N
Comment in case any doubt please rate my answer...
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