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 0.24. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 24.0kg and moment of inertia 0.500 kg?m2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.400m from that axis.

a) What is the acceleration of the block down the plane?

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 0.24. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 24.0kg and moment of inertia 0.500 kg?m2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.400m from that axis.

Explanation / Answer

Newton's 2nd law, for m , is:
m g + T + R = m a (vectors)
m g sen36.9

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