匿名
匿名 發問於 科學及數學其他 - 科學 · 7 年前

求高手 幫忙 摩擦力的問題?

http://a.imageshack.us/img843/6755/3pjs.jpg

摩擦力的問題 有些困難 謝謝幫忙喔^^

2 個解答

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  • 天同
    Lv 7
    7 年前
    最愛解答

    Consider the 10 kg slab alone, use: net-force = mass x acceleration

    100 - Ff = 10a ---------------- (1)

    where Ff is the friction between the block and slab

    a is the acceleration of the slab

    But Ff = u(10g)

    Here, 10g is the weight of the block

    u is the coefficient of friction (摩擦系數)

    Thus, (1) becomes,

    100 - 10ug = 10a

    a = 10 - u.g ---------------- (2)

    If the block goes with the slab without any sliding, the block and slab have the same acceleration a.

    Hence, 100 = (10 + 10)a

    a = 100/20 m/s^2 = 5 m/s^2

    Using (2): 5 = 10 - u(9.81)

    u = (10-5)/9.81 = 0.51

    Therefore, if u = 0.51, the block would have the same acceleration of 5 m/s^2 as the slab.

    If u < 0.51, from (2), the acceleration of the slab would increase. When u ~ 0, the acceleration of the slab = 100/10 m/s^2 = 10 m/s^2

    Hence, the acceleration of the slab is between 5 ~ 10 m/s^2

    The acceleration of the block a' is

    a' = Ff/10 = u10g/10 = ug

    Thus, 0 < a' < 5 m/s^2

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