1.民航客機機艙緊急出口的氣囊是條連接出口與地面的斜面,若斜面高3.2m, 斜面長5.5m,質量60kg 的人沿斜面滑下時所受的阻力是240N,求人滑至底端時的速度
2.質量500G的球被踢出後,某人觀察它在空中的飛行情況，估計上升的最大高度是10m，在最高點的速度為20m/s。請你根據這個估計，計算運動員踢球時對足球做的功。
高中物理 動能定理
2009-09-29 4:23 am
回答 (2)
2009-09-29 4:55 am
✔ 最佳答案
1. Potential energy at top of slope = 60g.(3.2) J = 1920 Jwhere g is the acceleration due to gravity (= 10 m/s2)
Work done by friction = 240 x 5.5 J = 1320 J
Hence, by conservation of energy,
potential energy at top of slope = kinetic energy at bottom + work-done against friction
1920 = (1/2)(60)v^2 + 1320
i.e. 30v^2 = 600
v^2 = 20 (m/s)^2
v = square-root[20] m/s
2. Potential energy at the highest point = (0.5)g(10) J = 50 J
Kinetic energy at highest point = (1/2)(0.5)(20)^2 J = 100 J
Hence, work done on the ball = (50+100) J = 150 J
2009-10-03 4:39 am
Yes la, you are right
收錄日期: 2021-04-29 17:31:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090928000051KK01385