香港寵物小精靈村落 論壇

標題: phy問題=) [打印本頁]
作者: 小霞的fans    時間: 3/9/2009 04:22 PM
標題: phy問題=)
本帖最後由 小霞的fans 於 2/9/2009 08:48 PM 編輯

assume air resistance is negligable.a ball is thrown veically upward with velocity v and it returns to its original position after time t.velocity and acceleration pointing upwards is taken to be positive.which of following is/are correct?
1.its velocity is zero at 0.5t.
2.the acceleration is maxium at t=0
3.at t seconds,its velocity is -v
我知2 一定錯..但係唔明3點解岩亦都唔知1岩唔岩..望指教..
另外..我有相似既問題..

assume air resistance is negligable.a ball is thrown veically upward with velocity 30ms^-1 . acceleration pointing downwards is taken to be positive.find the distance travelled after 1 second.
咁姐係u=30ms^-1 ,,a=10ms^-2,,t=1 s=?
s=ut+1/2at^2
s=30+1/2(-10)
=25m

但係佢既velocity唔係應該係20ms^-1(比acceleration每秒鍾-10米)架咩?
咁應該果秒行20米架喎,,點解會係25既-0- thx!!

P=Fv
'F'is the applied force that opposes resisting force acting on the object moving at a constant speed.
呢句野姐係話個F 係MOTION既相反方句!? opposes 同 resisting都解妨礙-0- , , 所以有d唔明
作者: sy    時間: 3/9/2009 04:48 PM
但係佢既velocity唔係應該係20ms^-1(比acceleration每秒鍾-10米)架咩?
>>這些公式好用於不用考慮這種東西,代入數字即可
作者: 超夢夢的夢境    時間: 3/9/2009 09:00 PM
assume air resistance is negligable.a ball is thrown veically upward with velocity v and it returns to its original position after time t.velocity and acceleration pointing upwards is taken to be positive.which of following is/are correct?
小霞的fans 發表於 3/9/2009 04:22 PM

                               
登錄/註冊後可看大圖

清楚起見,下面v改用u
代式
v^2-u^2=2as
it returns to its original position after time t
=> s=0

v^2=u^2
v=u or v=-u

v=u 係 initial velocity
所以v=-u

清楚D就係
v=u+at
a=/=0 and t=/=0
=> v=/=u
所以v=-u

v=-u and v=u+at
-u=u+at
-2u=at
-u=a(t/2)

(v at time=t/2)=u+a(t/2)=0

1,3

assume air resistance is negligable.a ball is thrown veically upward with velocity 30ms^-1 .acceleration pointing downwards is taken to be positive.find the distance travelled after 1 second.
小霞的fans 發表於 3/9/2009 04:22 PM

                               
登錄/註冊後可看大圖

u=-30ms^-1, a=10ms^-2, t=1
s=ut+1/2*at^2
s=-30+1/2*(10)
=-25m
(上行25米)

但係佢既velocity唔係應該係20ms^-1(比acceleration每秒鍾-10米)架咩?
咁應該果秒行20米架喎,,點解會係25既-0- thx!!
小霞的fans 發表於 3/9/2009 04:22 PM

                               
登錄/註冊後可看大圖

velocity at time =1s 係 20ms^-1...
但係佢唔係uniform velocity...
initial velocity = 30 ms^-1
因為 acceleration 係 uniform,
average velocity = (v+u)/2 =25

P=Fv
'F'is the applied force that opposes resisting force acting on the object moving at a constant speed.
呢句野姐係話個F 係MOTION既相反方句!? opposes 同 resisting都解妨礙-0- , , 所以有d唔明
小霞的fans 發表於 3/9/2009 04:22 PM

                               
登錄/註冊後可看大圖

applied force thatopposesresisting force即係同resisting force相反方向
In order words,in the same direction as motion

p.s."moving at a constant speed" implies applied force = resisting force
作者: 小霞的fans    時間: 5/9/2009 03:58 PM
本帖最後由 小霞的fans 於 4/9/2009 08:01 PM 編輯
清楚起見,下面v改用u
代式
v^2-u^2=2as
it returns to its original position after time t
=> s=0

v^2=u^2
v=u or v=-u

v=u 係 initial velocity
所以v=-u

清楚D就係
v=u+at
a=/=0 and t=/=0
= ...
超夢夢的夢境 發表於 3/9/2009 01:00 AM

                               
登錄/註冊後可看大圖


我明啦,, 好多謝你幫左我咁多^^ 仲有1個關於action and reaction pair既問題..

two block A and B are in contact.a force of 20N pushes the block fron the left

大約幅圖係咁..


                               
登錄/註冊後可看大圖


LET F(A) be the force on B by A. LET F(B) be the force on A by B

20=3+5(a)
a=2.5ms^-2

force acting on A by B
條式唔係
20+F(A)-F(B)=3a咩!?
20同F(A)係同方向既force (向右,,而F(B)係向左既,,
唔係右加右減左咩 .. .
點解會係20-F(A)=3a架!?同方向都相減!? 定係我抄錯,,唔係減F(A)係減F(B)..咁F(A)唔駛理!?個物件 exert 既force唔會計!?因為我睇返本書之後唔明 thx ><
作者: 捕風雪影    時間: 6/9/2009 01:22 PM
本帖最後由 捕風雪影 於 6/9/2009 01:23 PM 編輯
我明啦,, 好多謝你幫左我咁多^^ 仲有1個關於action and reaction pair既問題..

two block A and B are in contact.a force of 20N pushes the block fron the left

大約幅圖係咁..

http://i554.photobu ...
小霞的fans 發表於 5/9/2009 03:58 PM

                               
登錄/註冊後可看大圖


其實佢地係Action and Reaction Pair,所以Acting on A by B同埋Acting on B by A既force係相同的

所以A受到既Net Force=7.5N,咁Acting on B by A=7.5N,反之亦然
作者: 小霞的fans    時間: 6/9/2009 02:46 PM
其實佢地係Action and Reaction Pair,所以Acting on A by B同埋Acting on B by A既force係相同的

所以A受到既Net Force=7.5N,咁Acting on B by A=7.5N,反之亦然
捕風雪影 發表於 5/9/2009 05:22 PM

                               
登錄/註冊後可看大圖


但係acting on A by B同acting on B by A 既force都係12.5N.
我唔係好明呢d數點計- -  "
阿sir 將block  既mass(5)x a (2.5)=12.5N咁就計完. .
咁個20N 唔駛理架咩-,- .. 佢有份受力>< " 唔明呀!!!!!!
作者: fish    時間: 6/9/2009 03:03 PM
本帖最後由 fish 於 6/9/2009 03:10 PM 編輯
但係acting on A by B同acting on B by A 既force都係12.5N.
我唔係好明呢d數點計- -  "
阿sir 將block  既mass(5)x a (2.5)=12.5N咁就計完. .
咁個20N 唔駛理架咩-,- .. 佢有份受力>< " 唔明呀!!!!!!
小霞的fans 發表於 6/9/2009 14:46

                               
登錄/註冊後可看大圖


用free body diagram 來解釋不知會否好些呢?(看下圖, 只show horizontal component, no friction, right = positive, left = negative)

                               
登錄/註冊後可看大圖

即使兩個物體是放在一起時,那20N也只會畫在3kg上, 因為那20N是直接加於3kg上

Force on A:
F(x net) = 20N + F(B)

ma = 20N + F(B)

3kg * 2.5 m/s^2  = 20N + F(B)

F(B) = 7.5N - 20N = -12.5N

當然....直接點

Force on B:
F(x net) = F(A)

ma = F(A)       (and we know F(A) = - F(B) from the third law)

5kg *  2.5m/s^2 = - F(B)

F(B) = -12.5N
作者: 小霞的fans    時間: 6/9/2009 03:30 PM
用free body diagram 來解釋不知會否好些呢?(看下圖, 只show horizontal component, no friction, right = positive, left = negative)

                               
登錄/註冊後可看大圖

即使 ...
fish 發表於 5/9/2009 07:03 PM

                               
登錄/註冊後可看大圖


喔,,我大致都明了^^
Force on A:
F(x net) = 20N + F(B)

20N 同F(B)係相反方向點解係加既?
20N-F(B)=F(x net) 都得!?
作者: 捕風雪影    時間: 6/9/2009 03:34 PM
喔,,我大致都明了^^
Force on A:
F(x net) = 20N + F(B)

20N 同F(B)係相反方向點解係加既?
20N-F(B)=F(x net) 都得!?
小霞的fans 發表於 6/9/2009 03:30 PM

                               
登錄/註冊後可看大圖


正確寫法係

20N+(-F(B))=F(x net)

20N-F(B)=F(x net)

因為相反方向的
作者: fish    時間: 6/9/2009 03:40 PM
喔,,我大致都明了^^
Force on A:
F(x net) = 20N + F(B)

20N 同F(B)係相反方向點解係加既?
20N-F(B)=F(x net) 都得!?
小霞的fans 發表於 6/9/2009 15:30

                               
登錄/註冊後可看大圖

習慣問題....
我如果直接加的話在答案會出了個-12.5N (即是向左)

如果是  20N -F(B) =ma =7.5 的話
會出 F(B) = 12.5N (正數)
然後在report answer 時要寫 -12.5N 而已

不過我不清楚CE 是否都准許兩個方法了
作者: 捕風雪影    時間: 6/9/2009 03:42 PM
習慣問題....
我如果直接加的話在答案會出了個-12.5N (即是向左)

如果是  20N -F(B) =ma =7.5 的話
會出 F(B) = 12.5N (正數)
然後在report answer 時要寫 -12.5N 而已

不過我不清楚CE 是否都准許兩個方法 ...
fish 發表於 6/9/2009 03:40 PM

                               
登錄/註冊後可看大圖


CE接受兩種既答案

1.12.5N(Left)

2.-12.5N(但係你之前會指明邊一面係左,邊一面係右先可以用呢個答案)

Marker都會知道架喇
作者: 小霞的fans    時間: 7/9/2009 02:25 PM
我終於知我唔明咩..


                               
登錄/註冊後可看大圖


呢個係分開既free body diagram..

BLOCK A 既net force點解唔使理F(A)既,,BLOCK B既net force又唔理F(B)!?
佢自己exert架喎, ,,
一個block由斜台滑落黎都要理埋佢exert既mgsinθ 架喎
都要計埋佢自計到net force..
計block A 既net force 唔係應該係20+F(A)-F(B)=3X2.5ms^-2 咩!?
唔使加F(A)!? 唔係好明..
作者: fish    時間: 7/9/2009 02:51 PM
本帖最後由 fish 於 7/9/2009 02:53 PM 編輯
我終於知我唔明咩..



呢個係分開既free body diagram..

BLOCK A 既net force點解唔使理F(A)既,,BLOCK B既net force又唔理F(B)!?
佢自己e ...
小霞的fans 發表於 7/9/2009 14:25

                               
登錄/註冊後可看大圖


Free body diagram 只計/畫 force exert ON the body
自己exert的不是 exert ON itself, 而是 on others
所以 Force exertson Bby A 是不會在A 的 free body diagram中出現

我想你指的 mgsinθ是 將 weight 分做兩個components所得的其一
weight is a force exerts ON the block
因為於inclined plane時, weight跟其他的force (friction, normal force) 方向不一致
所以我們把weight 分做 mgsinθ 和 mgcosθ 方便計算而已
那不是block exerts force on something, 而是 gravity exerts forceon the block



歡迎光臨 香港寵物小精靈村落 論壇 (https://proxy.archiver.hkpnve.pokebeacon.com/)Powered by Discuz! X3.2