You attach a 1.70 kg block to a horizontal spring that is fixed at one end. You pull the block until the spring is stretched by 0.200 m and

Question

You attach a 1.70 kg block to a horizontal spring that is fixed at one end. You pull the block until the spring is stretched by 0.200 m and release it from rest. Assume the block slides on a horizontal surface with negligible friction. The block reaches a speed of zero again 0.200 s after release (for the first time after release). What is the maximum speed of the block (in m/s)

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Vodka 4 years 2021-08-22T09:03:19+00:00 1 Answers 21 views 0

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  1. Adela
    0
    2021-08-22T09:05:18+00:00
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    Answer:

    maximum speed of the block is 3.14 m/s

    Explanation:

    given data

    mass = 1.70 kg

    stretch in the spring = 0.2 m

    time take by block to come to zero  t = 0.2 s

    solution

    we know that Time period of oscillation (T) that is express as

    T = 2t    ………………….1

    put here value

    T = 2 (0.2)

    T = 0.4 s

    so here time period is express as

    T = 2\pi \sqrt{\frac{m}{k}}     …………….2

    here k is spring constant of the spring so  put here value

    0.4  =  2(\pi ) \sqrt{\frac{1.70}{k}}  

    here k will be

    k = 419.02 N/m

    so we use here conservation of energy that is

    Maximum kinetic energy = Maximum spring potential energy   …………3

    (0.5) m v² = (0.5) k x²

    here v is maximum speed block

    so put here value and we get

    (1.70) v² = (419.02) (0.2)²

    v = 3.14 m/s

    so maximum speed of the block is 3.14 m/s

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