1)將質量為M的木槐分成質量為m1 m2 兩陪分,並用細線連接,m1置於光滑的水平桌面上,m2通過定滑輪豎直懸掛,m1和m2有何種關係才能使系統在加速運動過程中繩的拉力最大?拉力的最大值是多少?
2)一水平放置的水管距地面高h=1.8m,水管橫截面積S=2cm^2.水從管口以不變的速度v=2m/s 連續沿水平方向射出,設出口橫截面上各處水的速度都相同,並假設水流在空中不散開,不計空氣阻力,求水流穩定後在空中水的體積.
物理難題(F4)(急要,謝謝)(20分)
2009-02-20 2:42 am
回答 (3)
2009-02-20 7:24 am
✔ 最佳答案
1.請問學左微分未?學左就睇下.F=ma
m2g=(m1+m2)a
a=m2g/(m1+m2)
T=m1a
T=m1m2g/(m1+m2)
T=m1(M-m1)g/(m1+M-m1)
T=m1g-gm12/M
dT/dm1=g-2gm1/M
for dT/dm1=0,
g-2gm1/M=0
m1=M/2
所以,當m1=m2,T是最大值.
跟據上述步驟,
T是最大值時:T=m1m2g/(m1+m2)
T=0.5M*0.5Mg/(0.5M+0.5M)
T=Mg/4
2.水柱在空在的時間:
s=ut+0.5gt^2
1.8=0.5(10)t^2
t=0.6s
水柱在空中的長度:
s=vt
s=2*0.6
s=1.2m
水柱的體積:
1.2*0.02=0.024m^3
2009-02-19 23:34:10 補充：
在水柱體積的計算中算錯了.
應該是,1.2*2*0.01*0.01=2.4*10^-4 m^3
參考: myself
2009-02-21 7:15 pm
謝謝兩位的解答
2009-02-20 5:13 am
1. Let T be the tension in the string, we have
For m1: T- m1.g = m1.a
where a is the acceleration of the system and g is the acceleration due to gravity
For m2: m2.g - T = m2.a
dividing: (T-m1.g)/(m2.g-T) = m1/m2
i.e. m2.T-m1.m2.g = m1.m2.g - m1.T
solve for T gives T = 2m1.m2.g/(m1+m2)
Let m1 = n. m2, where n is real number
T = 2.n.m2^2.g/m2(n+1)
T = 2.n.m2.g/(1+n)
Consider the fraction n/(1+n)
n/(1+n) = 1/(1+1/n)
In order for the fraction to be max, n should be very large
i.e. m1 should be larger compared with m2 for T to be maximum.
When n is large, 1/(1+1/n) approaches 1
therefore, T = 2.m2.g
or in terms of m1, T = 2(M-m1).g
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2. Forst find the time required for the water to go from the mouth of t he pipe to reach ground
Consider the vertical component of the water stream, using equation of motion,
u = 0 m/s, a = 1o m/s2, s = 1.8 m, t = >
use s = ut + (1/2).a.t^2
1.8 = (1/2)(10)t^2
t = 0.6 s
The amount of water leaving the pipe during 0.6 s
= 2 x 0.6 x (2x10^-4) m3 = 2.4x10-4 m3
This is the amount of water in air.
2009-02-19 21:34:30 補充：
Sorry...I made a mistake in Q1, the equation for m1 is T = m1.a only
The steps followed are similar, only that we would get T = m1.m2.g/(m1+m2)
The max T will then be T = m2.g in term of m2 or when in terms of m1,T = (M-m1).g
For m1: T- m1.g = m1.a
where a is the acceleration of the system and g is the acceleration due to gravity
For m2: m2.g - T = m2.a
dividing: (T-m1.g)/(m2.g-T) = m1/m2
i.e. m2.T-m1.m2.g = m1.m2.g - m1.T
solve for T gives T = 2m1.m2.g/(m1+m2)
Let m1 = n. m2, where n is real number
T = 2.n.m2^2.g/m2(n+1)
T = 2.n.m2.g/(1+n)
Consider the fraction n/(1+n)
n/(1+n) = 1/(1+1/n)
In order for the fraction to be max, n should be very large
i.e. m1 should be larger compared with m2 for T to be maximum.
When n is large, 1/(1+1/n) approaches 1
therefore, T = 2.m2.g
or in terms of m1, T = 2(M-m1).g
-------------------------------------------------------
2. Forst find the time required for the water to go from the mouth of t he pipe to reach ground
Consider the vertical component of the water stream, using equation of motion,
u = 0 m/s, a = 1o m/s2, s = 1.8 m, t = >
use s = ut + (1/2).a.t^2
1.8 = (1/2)(10)t^2
t = 0.6 s
The amount of water leaving the pipe during 0.6 s
= 2 x 0.6 x (2x10^-4) m3 = 2.4x10-4 m3
This is the amount of water in air.
2009-02-19 21:34:30 補充：
Sorry...I made a mistake in Q1, the equation for m1 is T = m1.a only
The steps followed are similar, only that we would get T = m1.m2.g/(m1+m2)
The max T will then be T = m2.g in term of m2 or when in terms of m1,T = (M-m1).g
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