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

Question

A block with mass m = 5.00 kg slides down a surface inclined 36.9 degree to the horizontal. 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. Part A What is the acceleration of the block down the plane? A = 2.5 m/s^2 Part B What is the tension in the string? T = 7.85 N

Explanation / Answer

On block perpendicular to incline ,

Fnet = N - mgcos36.9 = 0

N = mgcos36.9

along the plane ,

Fnet = mgsin36.9 - f - T = ma

and f = u.N = u.mgcos36.9

mgsin36.9 - T - u.mgcos36.9 = ma   .... (i)

On flywheel :

Net torque = r x T = I x alpha

alpha = a/r

T = I (a/r^2)

putting the value of T in eqtn (i) ,

5gsin36.9 - 0.500 x (a / 0.4^2) - 0.27 x5g x cos36.9 = 5a

18.86 = 8.125a

a = 2.32 m/s2 ....Ans (A)

B) T = I (a / r^2 )

T = 0.500 x 2.32 / 0.4^2 = 7.25 N

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