A block with mass 5.00kg slides down a surface inclined 36.9 degrees to the hori

Question

A block with mass 5.00kg slides down a surface inclined 36.9 degrees to the horizontal. The coefficient of friction is 0.26. A string attached to the block is wrapped around a flyhweel on a fixed axis at O. The flywheel has mass 22.0kg and the moment of inertia is .500kg*m^2 with respect to the axis of orientation. The string pulls without resistance at a perpendicular distance of .400m from that axis.

Part A.

What is the accleration of the block down the plane?

Part B.

What is the tension in the string?

A block with mass 5.00kg slides down a surface inclined 36.9 degrees to the horizontal. The coefficient of friction is 0.26. A string attached to the block is wrapped around a fly wheel on a fixed axis at O. The flywheel has mass 22.0kg and the moment of inertia is .500kg*m^2 with respect to the axis of orientation. The string pulls without resistance at a perpendicular distance of .400m from that axis.

Explanation / Answer

Mass of block M = 5 kg

Angle of incline A = 36.9 degrees

Coefficient of friction u = 0.26

Moment of inertia of wheel I = 0.5 kg-m2

Distance of string from wheel axis R = 0.4 m

Forces acting on the block are Mg downwards, tenstion T, normal reaction N (perpendicular to the surface of the incline) and friction f (along the direction of the incline, directed upwards), as shown below:

Let the acceleration of the block be a, directed downwards along the incline. Then by force balance, we have:

N = Mg*cosA

Mg*sinA - T - f = Ma => Mg*sinA - T - uMg*cosA = Ma

Also, the same tenstion T acts on the wheel and provides torque = TR. Angular acceleration of wheel is a/R.

So, TR = I * (a/R) => a = T*(R^2)/I

Then, MgsinA - T - uMgcosA = MT(R^2)/I => T = Mg(sinA - ucosA)/{M*(R^2)/I + 1} = 7.4 N

So, a = 7.4*(0.4*0.4)/0.5 = 2.368 m/s2

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