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 0.27. 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 kg?m2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.300m from that axis.

What is the acceleration of the block down the plane?

What is the tension in the string?

Explanation / Answer

PART A

Balancing forces:

m*g*sin(36.9) - T - 0.27*m*g*cos(36.9)= m*a --------------(1)

Taking torque of flywheel:

T*r = I a/r ---> T = (I/r^2) a

Substituting in (1)
m g(sin36.9

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