一質量為5kg的物體正沿著一斜面下滑,而該斜面和水平面成30度之夾角。
如果該物體和斜面間之摩擦力為20N,斜面長3m,求該物體離開斜面時之速度。
我知道可以用能量守恆原理,但加左摩擦力就唔識寫條式喇....唔該幫幫忙,plx!!^_^
緊急!!! 力學題目!! 唔該幫幫手~~ thx
2007-03-04 9:50 pm
回答 (3)
2007-03-04 10:21 pm
✔ 最佳答案
想問一問物體是否從靜止開始下滑如果不是，這題大概是計不了的
這題的確可以用能量守恆來計
為方便講解，定A點為斜面最高面，B為離去點
重力加速度即定為10(會考程度一般定為10較為準確應是9.81)
斜面高=3*sin30=1.5
物體在A點總能量 - 摩擦力對物體所作的功=物體在B點的總能量
(A點的勢能+A點的動能) - (摩擦力對物體所作的功)=B點的動能
註:由於沒有初速度，故A點動能為0
(5*10*1.5) - (20*3)=1/2*5*v平方 (註:v為離開斜面之速度)
v =6平方根
=2.45ms-1(三位有效數字)
2007-03-04 10:21 pm
Assume the object starts from rest
Potential energy (PE) at top of the slope = m.g.L.sin(30)
where m is the mass of the object (=5 kg)
g is the acceleration due to gravity
L is the length of slope (=3m)
Kinetic energy (KE) at bottom of slope = (1/2)m.v^2
where v is the velocity at bottom of slope
Work done against friction Wf = Ff.L
where Ff is the frictional force (=20N)
By conservation of energy
PE = KE + Wf
i.e. m.g.L.sin(30) = (1/2)m.v^2 + Ff.L
Potential energy (PE) at top of the slope = m.g.L.sin(30)
where m is the mass of the object (=5 kg)
g is the acceleration due to gravity
L is the length of slope (=3m)
Kinetic energy (KE) at bottom of slope = (1/2)m.v^2
where v is the velocity at bottom of slope
Work done against friction Wf = Ff.L
where Ff is the frictional force (=20N)
By conservation of energy
PE = KE + Wf
i.e. m.g.L.sin(30) = (1/2)m.v^2 + Ff.L
2007-03-04 10:05 pm
F=力 F=ma
m=質量
a=加速度
s=距離
單位=N F=ms
如果冇記錯就係啦
2007-03-04 14:06:39 補充：
條式係F=ma,,,,,,F=ms
m=質量
a=加速度
s=距離
單位=N F=ms
如果冇記錯就係啦
2007-03-04 14:06:39 補充：
條式係F=ma,,,,,,F=ms
參考: me
收錄日期: 2021-04-29 17:22:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070304000051KK02132