1/2mv^2=mgh?

Q: 1/2mv^2=mgh?
by the conservation of energy
1/2mv^2=mgh成立
但幾時可以用得(我知energy conserved..但邊d motion)

A: Your concept is entirely wrong. There is no such equation as (1/2)m.v^2 = mgh.

The Law of Conservation of Mechanical Energy states that, for an isolated system, the sum of kinetic and potential energy remains constant.

Mathematically, (1/2)m.v^2 + mgh = constant
In other words, the change of kinetic energy = change of potential energy
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Q: 同埋
mgsinQ係inclined road上面係咪永遠都係個friction

A: The answer is clearly NO.

The term [mg.sinQ] only represents the weight component of an object along an inclined plane with angle Q.

1)1/2mv^2=mgh 千萬不要幾時都用!如果唔係就死硬!(我都試過一次)。總之Intial total energy=Final total energy la!
2)唔知你問乜。如果放塊物件係inclined road上，而冇acceleration既話，同埋冇其他力既話，咁mgsinQ就等於friction 。但係如果有個力將佢向上推，而物件冇accleration,這裡還有二種情況，第一種是物件的的mgsinQ大過f而no accleration,咁f+F=mgsinQ,第二種是F大過mgsinQ and f,而no accleration,咁F=mgsinQ+f

涉及高度同速度的轉變ge時候....如果冇energy loss (friction heat sound energy....)就可以用呢條式...係inclined plant 都用得..
通常都係可以用係過山車..跳水..跳高咁.
但有時又咪係淨係1/2mv^2=mgh...
又時係1/2mv^2=mgh+1/2mv^2
當然兩個velocity 是不同的了...
例如..一個ball係地下右邊以10ms^-1移動..then有個inclined road高10米..問個ball係離地2米時..個velocity係幾多..
假設no energy loss
1/2mv^2=mgh+1/2mv^2
v^2=2gh+v^2
10^2-2x10x2=v^2
v=開方60ms^-1
又以上面個題...問個ball由地下可以升到離地幾高..
1/2mv^2=mgh
(1/2)v^2=gh
h=(1/2)10^2/10
h=5m

而另外一題mgsinQ係inclined road上面係咪永遠都係個friction
個物件有冇移動先??條題目講得唔係咁清楚..再補充我再答

第1個
當1件object由下往上郁到停果下
全部K.E轉做P.E(當冇energy loss)

同樣 當1年物體由上往下郁到觸地果下
全部P.E轉做K.E

都可以用le條式

第2個
都可以咁講