physics question

2008-04-21 6:39 am
A block of mass 5 kg is moving horizontally on a smooth surface in a straight line at 6m/s. In order to change its speed to 10m/s,then what is the work done need to be?

回答 (6)

2008-04-21 7:03 am
✔ 最佳答案
Work done required = Gain in KE
= ( 1 / 2 )( m )( v2 - u2 )
= ( 1 / 2 )( 5 )( 102 - 62 )
= 160J

2008-04-21 11:15:22 補充:
錯的應該是你!
Gain in KE = New KE - Initial KE
= ( 1 / 2 )( mv^2 ) - ( 1 / 2 )( mu^2 )
= ( 1 / 2 )( m )( v^2 - u^2 )

2008-04-21 11:20:14 補充:
如果好似你咁計, 咁你點解釋點解到2vu lee個term先? 根本計KE = ( 1 / 2 )( mv^2 )
New KE 好自然係( 1 / 2 )( mv^2 ), 而Initial KE就係( 1 / 2 )( mu^2 ), Gain in KE 咪就係減咗咁得出嚟。

2008-04-21 11:25:50 補充:
如果你仍然堅持係40J嘅, 咁我可以計俾你睇:
40 = ( 1 / 2 )( 5 )( v^2 - 6^2 )
v = 7.21m/s < 10m/s
由此可見40J係錯嘅!

2008-04-21 11:33:11 補充:
真糟糕, maths都唔掂就走嚟答phy = =
唉......

2008-04-22 09:10:11 補充:
有啦, 你見唔到我用佢個答案再計一次計到 < 10m/s 證明佢錯咩^^
參考: My Phy Knowledge
2008-04-22 8:26 am
咁多勁人係到既?
我又入先@@
其實001你有冇諗過驗算既呢?
同埋睇下個ans同條問題合唔合邏輯...
咁就可以避免出錯既機會啦_~~

2008-04-22 19:42:43 補充:
梗係見到啦^_____^
不過我諗佢冇習慣做完check數
咁係好蝕底!
因為check數可以幫你拎番唔少分!
2008-04-22 3:38 am
其實有好多人都誤以為(v﹣u)^2是等於v^2﹣u^2
但ΔKE = (1/2)(m)(v^2)﹣(1/2)(m)(u^2)
ΔKE = (1/2)(m)(v^2﹣u^2)
而不是ΔKE = (1/2)(m)(v﹣u)^2
所以請注意這點!!!
2008-04-22 2:43 am
第二位,請你看清楚題目要求,唔好盲目咁答題。
2008-04-21 7:30 pm
Gabriella Montez 的答案正確.
首先, 第二位用錯了式, 以為:
(v - u)^2 = v^2 - u^2
其次, v - u 只係反映出 change in speed, 而非 change in KE, 知兩者係不同的.
2008-04-21 6:23 pm
work done=gain in KE
=0.5mv^2
=0.5(5)(10-6)^2
=2.5(16)
=40J

2008-04-21 10:29:32 補充:
樓上計錯數
work done=gain in KE
=0.5mv^2
不是=0.5m(v^2-u^2)
而是=0.5m(v-u)^2
所以=0.5m(v^2-2vu+u^2)
.......=0.5(5)(10^2-2(10)(6)+6^2)


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