# 力學(摩擦力)英文”非作業”題目

A small block of mass Mb=0.50kg is placed on a long slab of mass Ms=3.0kg as shown above. Initially, the slab is at rest and the block has a speed v0(初速度) of 4.0m/s to the right. The coefficient of kinetic friction between the block and the slab is 0.20, and there is no friction between the slab and the horizontal surface on which it moves.

1) Calculate vf(末速度)

2) Calculate the distance the slab has traveled at the moment it reaches

vf(末速度)

Rating

我的想法是：把這二物看成是一個系統，此系統與外界無摩擦力，即無能量進出，所以動量守恆。

就好像非完全彈性碰撞一樣，雖然因摩擦力有能量損失，但動量仍然守恆。

圖看不清楚。假設板車夠長，Mb會因摩擦力而停止在Ms上。因動量守恆

所以 Mb*V0 + Ms *0 = (Mb +Ms)Vf

0.5 * 4 + 3 * 0 = ( 0.5+3 ) Vf

Vf = 4 / 7 (m/s) → (板車及物)對地的速度..............(1)

摩擦力 f = μ(Mb) g = (Mb) (a物)

(a物) =μg = 0.2 * 9.8 = 1.96

Vf = V0 + (a物) t

4 / 7 = 4 + (-1.96) t

t = (24/7) /1.96 = 1.75 (s)

又摩擦力 f帶動板車移動

所以 f = μ(Mb) g = (Ms) (a車)

0.2 * 0.5 * 9.8 = 3 (a車)

(a車) = 0.327

位移S = 1/2(a車) t2

= 0.5 * 0.327 *1.752

= 0.5 (m)..................(2)

Source(s): 愛物理的黑手