shows two blocks connected by a cord (of negligible mass) that passes over a frictionless pulley (also of negligible mass). The arrangement

Question

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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Nick 5 years 2021-09-05T12:32:25+00:00 1 Answers 17 views 0

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  1. Tryphena
    0
    2021-09-05T12:34:20+00:00
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    Answer:

    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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