問

L.K.

唔知岩唔岩麻...[[tk_35]]
how about question 4[[tk_35]]

Zend

Q4, 我認為你應該識 =.=
actually 你只係要考慮R 同mg應該係同一面定係相反jei -.-

仲有你可以用某d 式 , e.g. v = sqrt 5gr , v = sqrt gr ne 類, 點用就你自己做啦

AL Physics , 最初我都讀得好差.... 當然後來都係差, 但點都夠應付考試啦

L.K.

我計唔到ans...[[tk_35]]
4.)a small object of mass 0.05 kg is released from rest at the rim of aheavy, smooth semi-spherical boel of radius 10cm.Find the force actingon the object by the bowl when it passes the bottom of the bowl
A.0.5N    B.1.0N
C.1.5N    D.2.0N


loss  PE=gain KE
0.05*10*0.1=1/2*0.05*v²
v²=2
F_net=mv²/r
F=0.05*2/0.1
  =1N...=.=

ans is 1.5...

[ 本帖最後由 L.K. 於 27-11-2008 19:22 編輯 ]

Zend

acting by the bowl = mg +mv^2/r = 1+0.5 = 1.5 lor

L.K.

點解....[[tk_15]] [[tk_35]] [[tk_23]]

Zend

N-mg = mv^2/r ma
係bowl底
個N 係個底比個力個object 到

N = mg+mv^2/r lor

L.K.

係唔係因為佢問acting on the object by the bowl
所以要+番個mg  ??

Zend

yes -.-

芒果熙∼＊

Q1.
要加大normal reaction唔係正係加大mg先得嫁嘛..
可以加定風翼  架車行 就好似d風撞埋架車到(relatively..)
咁d風就會exert左一個向下既force component落架車到
所以N=mg+F
N大左
架車行係靠friction
friction同N係正比既.....
所以N大左 friction大左
>>>Fc又大左  
F=mv^2/r        m同r係constant    所以當F上升 v就可以更大