What is the acceleration of the block down the plane? A block with mass m = 5.00
ID: 1905252 • Letter: W
Question
What is the acceleration of the block down the plane?
A block with mass m = 5.00kg slides down a surface inclined 36.9 ^circ to the horizontal (the figure. The coefficient of kinetic friction is 0.26. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 29.0kg and moment of inertia 0.500 {rm{ kg}}cdot{rm{m}}^2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.200m from that axis. What is the acceleration of the block down the plane?Explanation / Answer
irst, force due to friction is mew force normal. the force normal is the force due to gravity times cosine of theta (36.9degrees) force due to gravity is mg which is 9.81 times 5.00kg. so the force due to friction is mew times 9.81 times 5.00kg times cos(36.9). the force down the slope due to gravity is force due to gravity times sin(36.9), and is negative because i took the force due to friction as postive and they are in opposite directions. f = -mgsin(36.9) add them up and get the total force on the block disregarding the tension on the string. getting to that. There is a force acting on the pulley at the point of contact of the string. this is a torque equal to the magnitude of the force due to the block sliding down the plane times the distance away from the center of the fly wheel this torque is 4.71 kg m^2/ sec^2 the angular accleration is the torque over the moment of inertia, and the angular acceleration is also the actual acceleration at that radius over that radius. so the acceleration is the torque times the radius over the moment of inertia. this is attached to the block so the block can only accelerate at this linear speed, or else the string would stretch or break. so the acceleration of the block down the slide is 2.36 meters per second squared. the tension on the string is 18.86 kilogram meters per second squared.
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