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Both A and B

Masterinc Secure | https//session.masteringphysics.com/myct/mastering MasteringPhysics: Chapter 10 Assignment - Google Chrome Secure l https://session.masteringphy.cs.com/myct/ite Chapter 10 AssignmentProblem 10.64 ials Problem 10.64 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 29.0 kg and moment of inertia 0.500 kg m2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.350 m from that axis Figure 1 of 1 5.00 kg 36.9 pe here to search

Explanation / Answer

For block m

Fnet = m*a

m*a = m*g*sin - µ*m*g*cos - T

a = [g(sin - µcos) - T]/m

a = [9.81m/s² * (sin36.9º - 0.26*cos36.9º) - T]/5kg

a = [9.81m/s² * (0.6 - 0.208) - T]/5kg

a = 3.85m/s² - 0.2*Tkg

For the flywheel

= I*

without slipping = a/r

= 0.500kg-m2 * (3.85m/s2 - 0.2T kg) / 0.35 m

= 1.428571 kg-m * (3.85m/s² - 0.2T kg)

= 5.5 N-m - 0.285m*T

But we also have = T*r, so

T*0.350m = 5.5N-m - 0.285m*T

T*0.635m = 5.5N·m

T = 8.65 N

Part a:

a = 3.85m/s2 – (0.2*8.66)

a = 2.12 m/s²

Part b:

T = 8.65 N

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