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1 September, 13:16

shows two blocks connected by a cord (of negligible mass) that passes over a frictionless pulley (also of negligible mass). The arrangement is known as Atwood's machine. One block has mass m1 = 1.30 kg; the other has mass m2 = 2.80 kg. What are (a) the magnitude of the blocks' acceleration and (b) the tension in the cord?

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  1. 1 September, 15:40
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    a = 3.585 m/s^2

    T = 17.4 N

    Explanation:

    Given:

    - The mass of block 1 m1 = 1.30 kg

    - The mass of block 2 m2 = 2.8 kg

    - The acceleration of block-block system = a

    - The tension in the cord = T

    Find:

    (a) the magnitude of the blocks' acceleration

    (b) the tension in the cord?

    Solution:

    - In our coordinate systems, the acceleration of the block 1 equals the acceleration of the block 2. So we can express them as a common acceleration a. The equations of motion are:

    m1*a = T - m1*g

    m2*a = m2*g - T

    - Adding the two equations together, we get the equation of motion for the whole system.

    (m1+m2) * a = m2*g-m1*g

    - Solving this equation for the acceleration, we have

    a = (m2-m1) * g / (m1+m2)

    a = (2.8-1.3) * 9.8 / (1.3+2.8)

    a = 3.585 m/s^2

    - Plugging the result into the equation of motion for block 1, we have

    T = m1 * (a+g)

    T = 1.3 * (3.585+9.8)

    T = 17.4 N
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