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 0.24. A string attached to the block is wrapped around a flywheel on a fixed axis at O . The flywheel has mass 8.16 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 0.350 m from that axis.

Part A

What is the acceleration of the block down the plane?

Part B

What is the tension in the string? How do I approach and solve this problem as well as state the correct answer

Explanation / Answer

Let  the acceleration of the block down the plane, = a
Let the tension in the string, = T

r = 0.350

We know,
t(torque) = I *
Force * Perpendicular distance = I * (a/r)
T*r = I*(a/r)
T*r^2 = I*a
T*0.35^2 = 0.5*a
T = 4.08 * a ---------1

m*g*sin(36.9) - uk*m*g*cos(36.9) - T = m*a
5.0*9.8*sin(36.9) - 0.24*5.0*9.8*cos(36.9) - T = 5.0*a
20.0 - T = 5.0 *a ------2

Solving eq 1 & 2,
a = 2.20 m/s^2
T = 9.0 N

(a)
Acceleration of the block down the plane, a = 2.20 m/s^2
(b)
Tension in the string, T = 9.0 N

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