A 2.8 kg block slides with a speed of 2.4 m/s on a frictionless horizontal surface until it encounters a spring. Part A If the block compres

Question

A 2.8 kg block slides with a speed of 2.4 m/s on a frictionless horizontal surface until it encounters a spring. Part A If the block compresses the spring 5.6 cm before coming to rest, what is the force constant of the spring

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Khoii Minh 4 years 2021-08-26T06:46:16+00:00 1 Answers 25 views 0

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  1. thongdat2
    0
    2021-08-26T06:47:40+00:00
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    Answer:

    5,142.86

    Explanation:

    The kinetic energy possessed by the block when sliding will be equal to the energy needed to compress the string.

    Kinetic energy = 1/2 mv² and energy stored in the spring = 1/2 ke²

    m = mass of the block (in kg) = 2.8 kg

    v = speed of the block (in m/s) = 2.4 m/s

    k = force constant of the spring

    e = extension (in metres) = 0.056m

    Since KE = energy stored in the spring

    1/2 mv² = 1/2 ke²

    mv² = ke²

    2.8(2.4)²  = k(0.056)²

    16.128 = 0.003136k

    k = 16.128/0.003136

    k =  5,142.86

    The force constant of the spring is 5,236.36

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