Frictiona and tension problem
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Frictiona and tension problem

[From: ] [author: ] [Date: 12-02-18] [Hit: ]
67 * m2 =201.m2 = 75.........
In the following figure m1 = 20.0 kg and a = 50.7 degrees. The coefficient of kinetic friction between the block and the incline is mu k = 0.40.

What must be the mass m2 of the hanging block if it is to descend 12.0 m in the first 3.00 s after the system is released from rest?

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For block2 with mass m2:

3 forces: -mg, T2 and -ma

y = vt - ½at²

12 = 0 - ½a(3)²

Net acceleration = a = -2.67 m/s²
T = m2a = -2.67 * m2


For block1 with mass m1:

3 forces: Ff, m1gsinθ, T

T - Ff - m1gsinθ = 0

m2a - μm1gcosθ - m1gsinθ = 0

2.67 * m2 = (0.40)(20.0)(9.81)(cos 50.7º) + (20.0)(9.81)(sin 50.7º)

2.67 * m2 = 49.708 + 151.827

2.67 * m2 = 201.535

m2 = 75.48 kg

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