friction along inclined plane

When the block is moving up, it is subject to two retarding forces,
1. the weight component acting downward parallel to the inclide plane. The force is 1.5g.sin(20) N = 5.13 N (where g is the acceleration due to gravity)

2. the friction force Ff, which also acts downward along the plane.

Thus, 5.13 + Ff = 1.5 x 5 (net force = mass x acceleration)
this gives Ff = (1.5x5 - 5.13) N = 2.37 N
which is the result given in statement (2)
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Statement (1) is correct.
Normal reaction = 1.5g.cos(20) N

Statement (3) is not correct.
When the block comes to rest at the highest point, the weight component acting downward along the plane is 5.13 N

But frcitional force is only 2.37 N, which is insufficient to balance the downward weight component (notice that the friction is now acting upward along the plane, attempting to prevent the block from sliding down). The block thus begins to slide down.

why a= 5 but not -5

我會揀 ( A )

(1) normal reaction = mg cos 20
= 1.5(10)(cos 20)
=14.1N

(2)
F net = ma
mgsin20 + friction = 1.5(5)
friction = 2.36969785 N

Thus,我估岩

(3)錯,因為mgsin20 大過 frictional force

thus,有net force,所以去到max. pt.會趺返落黎=)

我諗係咁樣做掛 , 如果錯既就話返比我知啦=]

2008-06-16 22:28:05 補充：
因為 a= -5個負係指方向,

if block向右行係 take +ve

所以向右減速=向左加速 , thus, a= -ve

但係你個friction 同 mgsin20都係向左面施力,

then向左加力 就take左 -ve

所以你可以咁睇 :

-mgsin20-friction=1.5(-5)

-(mgsin20 + friction)=-7.5

mgsin20 + friction = 7.5

Thus, 最後都係一樣=]