A small block slides down an incline with a constant acceleration. The block is released from rest at the top of the incline. After it trave

Question

A small block slides down an incline with a constant acceleration. The block is released from rest at the top of the incline. After it travels 5 m to the bottom its speed is 5 m/s. What is the speed, in m/s, of the block when it had traveled only 2.1 m from the top

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Nho 2 months 2021-08-17T02:22:36+00:00 1 Answers 0 views 0

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    2021-08-17T02:24:06+00:00

    Answer:

    v = 3.24 m/s

    Explanation:

    Since we don’t have time, we can use the formula;

    (Final distance – initial distance)/time = (initial velocity + final velocity)/2

    Thus;

    (x_f – x_i)/t = ½(v_xi + v_xf)

    We are given;

    x_i = 0 m

    x_f = 5 m

    v_xi = 0 m/s

    v_xf = 5 m/s

    Thus, plugging in the relevant values;

    (5 – 0)/t = (0 + 5)/2

    5/t = 5/2

    t = 2 s

    Using Newton’s first law of motion, we can find the acceleration.

    v = u + at

    Applying to this question;

    5 = 0 + a(2)

    5 = 2a

    a = 5/2

    a = 2.5 m/s²

    To get the speed, in m/s, of the block when it had traveled only 2.1 m from the top, we will use the formula;

    v² = u² + 2as

    v² = 0² + 2(2.5 × 2.1)

    v² = 10.5

    v = √10.5

    v = 3.24 m/s

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