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Bess [88]
3 years ago
9

What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.16?

Physics
1 answer:
Gekata [30.6K]3 years ago
6 0
(Missing figure is here: https://www.physicsforums.com/attachments/ch05-p070-jpg.149243/ )

Let's call m_1 = 1.0 kg and m_2=2.0 kg the masses of the two blocks. We can write Newton's second law for both blocks (sum of all forces acting on the block = ma). On block 1, we have two forces: the weight m_1 g pointing downwards and the tension of the string T poiting upwards. On block 2, we have the tension of the string going right and the friction \mu m_2 g going left. Therefore

m_1 g - T = m_1 a
T-\mu m_2 g = m_2 a

Summing the two equations, we find
m_1 g - \mu m_2 g = (m_1 + m_2) a
and then using \mu=0.16 we can find the acceleration:
a= \frac{m_1 g - \mu m_2 g}{m_1 + m_2}=2.22 m/s^2


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Given:

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Now, putting the given values in the above formula:

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\alpha = 1.86 m/s^{2}

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You are given three resistors with the following resistances: R1 = 6.32 Ω, R2 = 8.13 Ω, and R3 = 2.29 Ω. What is the largest equ
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Let the time for the green car to catch up with the blue car be T

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