標題:
求高手 幫忙 摩擦力的問題?
http://a.imageshack.us/img843/6755/3pjs.jpg 摩擦力的問題 有些困難 謝謝幫忙喔^^
最佳解答:
Consider the 10 kg slab alone, use: net-force = mass x acceleration 100 - Ff = 10a ---------------- (1) where Ff is the friction between the block and slab a is the acceleration of the slab But Ff = u(10g) Here, 10g is the weight of the block u is the coefficient of friction (摩擦系數) Thus, (1) becomes, 100 - 10ug = 10a a = 10 - u.g ---------------- (2) If the block goes with the slab without any sliding, the block and slab have the same acceleration a. Hence, 100 = (10 + 10)a a = 100/20 m/s^2 = 5 m/s^2 Using (2): 5 = 10 - u(9.81) u = (10-5)/9.81 = 0.51 Therefore, if u = 0.51, the block would have the same acceleration of 5 m/s^2 as the slab. If u < 0.51, from (2), the acceleration of the slab would increase. When u ~ 0, the acceleration of the slab = 100/10 m/s^2 = 10 m/s^2 Hence, the acceleration of the slab is between 5 ~ 10 m/s^2 The acceleration of the block a' is a' = Ff/10 = u10g/10 = ug Thus, 0 < a' < 5 m/s^2
其他解答:
以下是我找到的資料 可以讓你參考看看 資料 http://disp.cc/b/518-3wyP http://www.xuexifangfa.com/physics/jieti/4234.html http://www.wenkudaquan.com/doc/20121106/887296.html http://dmt.zjlsedu.org/longresgz/300/040/120/070/L000000000092403.htm http://www.wuliwang.net/a/wulijiajiaoyanjiu/311.html 希望以上的資料對你會有幫助 感恩!!
求高手 幫忙 摩擦力的問題?
此文章來自奇摩知識+如有不便請留言告知
發問:http://a.imageshack.us/img843/6755/3pjs.jpg 摩擦力的問題 有些困難 謝謝幫忙喔^^
最佳解答:
Consider the 10 kg slab alone, use: net-force = mass x acceleration 100 - Ff = 10a ---------------- (1) where Ff is the friction between the block and slab a is the acceleration of the slab But Ff = u(10g) Here, 10g is the weight of the block u is the coefficient of friction (摩擦系數) Thus, (1) becomes, 100 - 10ug = 10a a = 10 - u.g ---------------- (2) If the block goes with the slab without any sliding, the block and slab have the same acceleration a. Hence, 100 = (10 + 10)a a = 100/20 m/s^2 = 5 m/s^2 Using (2): 5 = 10 - u(9.81) u = (10-5)/9.81 = 0.51 Therefore, if u = 0.51, the block would have the same acceleration of 5 m/s^2 as the slab. If u < 0.51, from (2), the acceleration of the slab would increase. When u ~ 0, the acceleration of the slab = 100/10 m/s^2 = 10 m/s^2 Hence, the acceleration of the slab is between 5 ~ 10 m/s^2 The acceleration of the block a' is a' = Ff/10 = u10g/10 = ug Thus, 0 < a' < 5 m/s^2
其他解答:
以下是我找到的資料 可以讓你參考看看 資料 http://disp.cc/b/518-3wyP http://www.xuexifangfa.com/physics/jieti/4234.html http://www.wenkudaquan.com/doc/20121106/887296.html http://dmt.zjlsedu.org/longresgz/300/040/120/070/L000000000092403.htm http://www.wuliwang.net/a/wulijiajiaoyanjiu/311.html 希望以上的資料對你會有幫助 感恩!!
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