A lead ball of mass 10kg is hung by a rope of length 2.0m The ball is pulled
aside for a vertical distance of 0.8m and is then released as shown in the
picture. Determine the horizontal distance form its orginal position before
the release and the velocity that the lead ball will pass its lowest position
after being released
圖片參考：http://i56.photobucket.com/albums/g175/chengqoo2000/leadball.jpg
lead ball problem
2011-01-13 5:04 am
回答 (1)
2011-01-13 6:23 pm
✔ 最佳答案
Vertical distance from the point of suspension= (2 - 0.8) m = 1.2 m
Hence, horizontal distance moved by the ball
=square-root[2^2 - 1.2^2] m =1.6 m
Using conservation of mechanical energy,mg(0.8) = (1/2)mv^2where m is the mass of the ball, g is the acceleration due to gravity, v is the speed of the ball at the lowest positionv = square-root[2g(0.8)] m/s = 4 m/s[taking g = 10 m/s2]
收錄日期: 2021-04-29 17:37:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110112000051KK01172