[F.4 phy]力學問題

1. Since the two boxes are in contact, they moves together when an applied force of 120 N is given, and would have the same acceleration.

Acceleration = net-force/mass = 120/(80+40) m/s^2 = 1 m/s^2

If you want to find the froce that acts on box B, the
Force = mass x acceleration = 40 x 1 N = 40 N

2. The man in the lift experiences an increase in weight when the lift accelerates upward, and then his weight returns to normal when the lift attains constant velocity.

The apparent weight R is given by
R = mg + ma
where m is the man's mass, g is the acceleration due to gravity, a is the acceleration of the lift.

Hence, R > mg (the man's normal weight) when a is not zero, and R = mg when a is zero (i.e. constant velocity)

Q: 因為如果要令lift由靜止既情況轉為均速上升,我認為一定要有力令lift moves upward,而所施既力會導致normal reaction會大於weight,導致apparent weight增加。而之後由於均速上升既關係意味著冇net force施加於lift,所以後期lift內既人就唔會感受到自己既apparent weight既改變...請問咁諗正唔正確?

A: No. You should consider the forces that acts on the man, not those on the lift.

1)
f=ma,
120N=(Box A + Box B)*a
120=(40+80)*a
120=120a
a=1ms^-2

因為兩個box互相緊扣,所以accleration一樣
比例是1:1, 加速度也是1ms^-2

2)
不會, 因為題中說明了是constant velocity, constant velocity 是指 a=0ms^-2
所以我們是不會感到apparent weight 的增加

有點不太正確
因為normal reaction = weight
在加速上升及at rest 時也一樣
但在加速上升時會多了一個力 叫 tension
tension 的力會令lift 加速及上升
所以是tension導致apparent weight增加, 和normail force 無關
之後的都對!! ^.^