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

Question

A block with mass m = 5.00 kg slides down a surface inclined 36.9 to the horizontal (See Fig 10.55 University Phsics 12Th). The coefficient of kinetic friction is 0.22. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 27.5 kg and moment of inertia 0.496 kg · m2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.190 m from that axis.

(a) What is the acceleration of the block down the plane?
(b) What is the tension in the string?
Thanks

Explanation / Answer

Given: Mass of the block = m = 5 kg Inclination angle = = 36.9o coefficient of kinetic friction is=k = 0.22 mass of the fly wheel = M = 27.5 kg moment of inertia of the flywheel = I = 0.496 kg · m2 perpendicular distance of = R = 0.190 m from that axis (a)   If T be the tension in the string , Acc. to given problem From free body diagram we have , mg (sin - k cos ) - T = ma --(1) where a is the acceleration of the block but , angualr acceleration;    = a / R Acc. to torque euqtion , we ahve = I but , = T R so, TR = I a/R      T = Ia /R2 --(2) sub.T from eqn: 2 in 1 we have , a = mg (sin -  k cos ) / ( m + I/R2 )      = (5)(9.8) (sin 36.9 - 0.22 cos36.9) / (5+ 0.496 /0.192)      =  49 (0.425) /18.739      = 1.111 m/s2 (b) but , from eqn: (2) we have , Tension in the string :     T = I a2 / R         = (0.496)(1.111)2 /(0.19)         = 3.222 N coefficient of kinetic friction is=k = 0.22 mass of the fly wheel = M = 27.5 kg moment of inertia of the flywheel = I = 0.496 kg · m2 perpendicular distance of = R = 0.190 m from that axis (a)   If T be the tension in the string , Acc. to given problem From free body diagram we have , mg (sin - k cos ) - T = ma --(1) where a is the acceleration of the block but , angualr acceleration;    = a / R Acc. to torque euqtion , we ahve = I but , = T R so, TR = I a/R      T = Ia /R2 --(2) sub.T from eqn: 2 in 1 we have , a = mg (sin -  k cos ) / ( m + I/R2 )      = (5)(9.8) (sin 36.9 - 0.22 cos36.9) / (5+ 0.496 /0.192)      =  49 (0.425) /18.739      = 1.111 m/s2 (b) but , from eqn: (2) we have , Tension in the string :     T = I a2 / R         = (0.496)(1.111)2 /(0.19)         = 3.222 N

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