# 求高手 幫忙 摩擦力的問題?

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### 2 個解答

• 天同
Lv 7
8 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