PLEASE HELP THIS PROBLEM IS DUE
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PLEASE HELP THIS PROBLEM IS DUE

[From: ] [author: ] [Date: 12-09-23] [Hit: ]
mA = (10 kg)[sin 30 - (0.mA = 3.......
Two blocks are connected over a pulley. The mass of block A is 10 kg, and the coefficient of kinetic friction between A and the incline is 0.20. Angle θ is 30°. Block A slides down the incline at constant speed. What is the mass of block B?

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Since A is moving down the incline,
the force of friction is uphill with magnitude
fk = μkFn

Since the block moves with constant speed, a = 0.
According to Newton's Second Law,
ΣFx = T + fk - mBg sinθ = 0 (1)
ΣFy = Fn - mBg cosθ = 0 (2)
mAg - T = 0 (3)

Solving (1) for T,
T = mBg sinθ - fk
or
T = mBg sinθ - μkmBg cosθ

Solving (3) for mA,
mA = T/g

Substituting for T,
mA = (mBg sinθ - μkmBg cosθ) / g
or
mA = mB sinθ - μkmB cosθ
or
mA = mB(sinθ - μk cosθ)

mA = (10 kg)[sin 30 - (0.2)cos 30]
mA = 3.3 kg
1
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