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 (the figure (Figure 1) ). The coefficient of kinetic friction is 0.26. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 16.0 kg and moment of inertia 0.500 kgm2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.250 m from that axis. A)What is the acceleration of the block down the plane?   B)What is the tension in the string?

Explanation / Answer

here,

mass of block , m = 5 kg
theta , A = 36.9 degrees
coefficient of kinetic friction, uk = 0.26
mass of flywheel = 16 kg
moment of inertia of flywheel, I = 0.5 kg.m^2
distance(r) = 0.250 m

Assuming:
Ff = frictional force = uk*Normal Force(N)
a = acceleration of block
T = tension in string
t = torque in flwheel
alpha = angular acceleration = a/r

From enwton second Law : Fnet = m * a
mgSin - T - Ff = m*a
mgSin - T - uk*m*g*cos = m*a
5*9.8*Sin36.9 - T - 0.26*5*9.8*Cos36.9 = 5*a
-T - 44 = 5a -------------------(1)

also from flywheel,
Tension due to string = torque in wheel
T*r = I*alpha
T*r = I * a/r
T = I*a/r^2 -------------------(2)

using 2 in 1 we get :

a = ( m * g(sin - cos) )/ (m + (I/r^2) )
a = ( 5 * 9.8(sin36.9 - 0.26*cos36.9) )/ (5 + (0.500/(0.250)^2) )
a = -1.479 m/s^2

Therefore eqn 2 becomes :

T = I*a/r^2
T = 0.5 * 1.479/(0.25)^2
T = 11.832 N

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