一0.777 kg質量、長度1.20 m和橫切面積8.30x 10-5m2的靜止圓筒在 2.30 m 的高度下墜,如撞擊地面為彈性撞擊及圓筒的收縮服從虎克定律並有1.60x10-3m的縮短，計算
1. 儲存在圓筒中最大的彈性勢能
2. 圓筒施加於地面的平均力
3. 圓筒施加於地面的力的最大值(ans : 21.9x10^3 N)
唔識計3 , 唔該解釋下
力學問題1 [15分] 急 !
2009-03-24 3:36 am
回答 (1)
2009-03-24 4:19 am
✔ 最佳答案
1. Loss of gravitational potential energy = gain in elastic potential energyhence, gain in elastic potential energy = 0.777 x 10 x 2.3 J = 17.871 J
2. use work done = change of kinetic energy
(F-0.777g) x1.6x10^-3 = 17.871
where F is the average retarding force
solve for F
3. The max force occured when the cylinder is fully compressed
since 17.871 = (1/2)k(1.6x10^-3)^2
where k is the elastic constant
k = (2x17.871)/(1.6x10^-3)^2 N/m
The max force = k(1.6x10^-3) + 0.777g N
k =
收錄日期: 2021-04-29 17:34:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090323000051KK01469