A block mass m = 5.00kg slides down a surface inclined 36.9 degrees to the horiz
ID: 1288213 • Letter: A
Question
A block mass m = 5.00kg slides down a surface inclined 36.9 degrees to the horizontal. The coefficient of kinetic friction is 0.27. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 29.0 kg and moment of intertia 0.500 kg*m^2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.300 m from that axis.
A. What is the acceleration of the block down the plane?
B. What is the tension in the string?
A block mass m = 5.00kg slides down a surface inclined 36.9 degrees to the horizontal. The coefficient of kinetic friction is 0.27. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 29.0 kg and moment of intertia 0.500 kg*m^2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.300 m from that axis. A. What is the acceleration of the block down the plane? B. What is the tension in the string?Explanation / Answer
A. On flywheel :
torque = rx F = I x alpha
0.300 x T = 0.500 x (a/0.3)
T = 5.56a
on block :
mgsin36.9 - u.mgcos36.9 - T= ma
5 x 9.8 x sin36.9 - 0.27 x 5 x9.8 x cos36.9 - T = 5a
18.84 - 5.56a = 5a
a = 1.78 m/s2
B. T = 5.56 x 1.78 = 9.92 N
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