An object,moving up a smooth inclined plane making an angle c with horizontal decreases its speed from x to y ms^-1.What is the distance,in metres,travelled in this period??(take g=acceleration due to gravity in ms^-2)
why the answer should be (x^2-Y^2)/2g sin c??
can i use v^2=u2+2as to solve the problem??
thx
力學問題...急...1974某條ce 快!!!
2011-05-31 6:03 am
回答 (2)
2011-05-31 6:57 am
✔ 最佳答案
Use equation of motion: v^2 = u^2 + 2a.swith u = x m/s, v = y m/s, a = -g.sin(c). s = ?
hence, y^2 = x^2 + 2.(-g.sin(c))s
s = (x^2 - y^2)/[2g.sin(c)]
2011-05-31 7:21 am
圖片參考：http://imgcld.yimg.com/8/n/HA00144763/o/701105300109913873453810.jpg
please look at the diagram,
if you resolve parallel to the plane (upward)Net force=ma-w sin c=ma -9.8m sin c=ma (cancel m from both side)A=-9.8 sin cA=-g sin c Since:U=xV=yA=-g sin cS=?T=/ We can use v^2=u^2+2asy^2=x^2+2(-g sin c)sx^2-y^2=(2g sin c)s∴s=x^2-y^2/2g sin chope it helps! =]
參考: myself
收錄日期: 2021-04-29 17:39:16
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