A broken-down car of mass 1000kg is pulled by a tow-truck and moves at a
constant velocity 8ms^-1 along a horizontal road.Lt is known that the
frictional force acting on the car is 500N.Find the tension in the cable
connecting the trunk and the car.
A.0N
B.500N
C.8000N
D.8500N
E.10500N
答係B，本人笨拙,無法理解
(hkcee 98)
1st statement : A car runs faster on a frictionless road.
答案話呢句子錯誤,但我都係唔明
一個盒左邊受了10N的力,右邊受了2N的力,It remain at rest on the table. If the force of magnitude 10N is removed,find the resultant force acting on the block.
A.zero
B.2N
C.6N
D.8N
E10N
答案係A
(hkcee 95)
完全唔明白:(
望各高人解答,小弟感激萬分
幾題力學入門
2008-04-27 3:51 am
回答 (3)
2008-04-29 4:01 am
✔ 最佳答案
(1)因為係constant velocity,所以net force=0so,T=f
since f=500N
T=500N
你第2題同第3題一樣既？？
(2),(3)當10N的力is removed,the block is remain at rest
so net force=0
f=2N
2-2=0N
resultant force=0N
notice:雖然仲有一個2N的力,resultant force仍等於0,因為friction是2N
你對net force既概念唔熟,有唔明既就盡快問清楚老師,要搞返清個概念
2008-04-27 8:10 pm
一個盒左邊受了10N的力,右邊受了2N的力,It remain at rest on the table. If the force of magnitude 10N is removed,find the resultant force acting on the block.
F(net)=8N
但佢都係at rest..即係話個friction可以去到8N
咁依家得番2N acting on佢身上...
NEWTON 3rd law of motion
action = reaction
friction=2N
F(net)=8N
但佢都係at rest..即係話個friction可以去到8N
咁依家得番2N acting on佢身上...
NEWTON 3rd law of motion
action = reaction
friction=2N
2008-04-27 4:28 am
第1題
你要計左 broken car 個net force先
咁條題目話moves at a constant velocity 即係 acceleration =0
take 左邊係+既話
F-friction=ma
F-500N=1000kg(0)
F=-500N
Tension 就係500N喇
第2題
A car runs faster on a frictionless road.
無friction 向唔到前 因為無action and reaction
第3題你係咪sure佢係95年@@?
2008-04-26 21:17:58 補充：
條題目係咪英文嫁?
你可唔可以將佢完整打番出黎, 唔好意思
你要計左 broken car 個net force先
咁條題目話moves at a constant velocity 即係 acceleration =0
take 左邊係+既話
F-friction=ma
F-500N=1000kg(0)
F=-500N
Tension 就係500N喇
第2題
A car runs faster on a frictionless road.
無friction 向唔到前 因為無action and reaction
第3題你係咪sure佢係95年@@?
2008-04-26 21:17:58 補充：
條題目係咪英文嫁?
你可唔可以將佢完整打番出黎, 唔好意思
參考: myself
收錄日期: 2021-04-16 12:49:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080426000051KK02572