A block with mass m = 5.00 kg slides down a surface inclined 36.9degree to the h

Question

A block with mass m = 5.00 kg slides down a surface inclined 36.9degree 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 6.25 kg 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 0.400 m from that axis. What is the acceleration of the block down the plane? 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 sen36.9° - T - fd m g cos36,9° = m a
T*r = I a/r ---> T = (I/r^2) a
Then
m g(sen36.9° - fd cos36.9°) = (m + I/r^2) a
a = {m g(sen36.9° - fd cos36.9°)/(m + I/r^2)

a = 5.00*9.81(sin36.9-0.27*0.400*cos36.9)/6.25+0.500/0.200

= 6.96 m/s^2
T = (0.5/0.4^2)*6.96 = 21.75N

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