2 個解答
- 天同Lv 78 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
