A block with mass m = 5.00 k g slides down a surface inclined 36.9 ? to the hori

Question

A block with mass m = 5.00kg slides down a surface inclined 36.9 ? to the horizontal (the figure (Figure 1) ). 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 23.0kg and moment of inertia 0.500 kg?m 2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.400m from that axis. Part A What is the acceleration of the block down the plane? Part B What is the tension in the string? A block with mass m = 5.00kg slides down a surface inclined 36.9 ? to the horizontal (the figure (Figure 1) ). 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 23.0kg and moment of inertia 0.500 kg?m 2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.400m from that axis. Part A What is the acceleration of the block down the plane? Part B What is the tension in the string? Part A What is the acceleration of the block down the plane? Part B What is the tension in the string? Part B What is the tension in the string?

Explanation / Answer

Newton's 2nd law, for m , is:

m g + T + R = m a (vectors)

m g sin36.9

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