Question

A small block with mass 0.0500 kg slides in a vertical circle of radius 0.0760 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

Part A: What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

HW 8_ SS19 (Chapters 7 and 8) Problem 7.71 4 of 10> Constants Part A A small block with mass 0.0500 kg slides in a vertical circle of radius 0.0760 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 What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path? Express your answer with the appropriate units magnitude 3.70 N Value Units SubmitPre vious Answers Request Answer xIncorrect Try Again Provide Feedback Next >

Answer 1

From above data that at bottom N - mg = centripetal force 3.70 - 0.0500 times 9.8 = mvA^2/r v_A = squareroot 3.21 times 0.0760/0.05 v_A = 2.24 m/s at top: Total energy at top = total energy at bottom mgh + 1/2 mv_B^2 = 1/2 mv_A^2 9.8 times 2 times 0.076 + 1/2 v_B^2 = 1/2 (2.21)^2 1.4896 + 1/2 v_B^2 = 2.44205 1/2 v_B^2 = 0.95245 v_B^2 = squareroot 1.9049 v_B = 1.38 m/s Now mg + N = centripetal force = 0.05 times 9.8 + N = 0.05 times (1.38)^2/0.076 N = 1.25289 - 0.49 N = 0.76 N