Two blocks are connected by a light string that passes over a frictionless pulley. The system is released from rest while m2 is on the floor and m1 is a distance h above the floor.
Taking m1=6.5kg, m2=4.2kg, and h=3.2m, what is the speed of m1 just as it reaches the floor?
phys pulley system
2013-12-20 7:46 am
回答 (2)
2013-12-20 9:01 am
✔ 最佳答案
Taking downward as +ve.net force on m1 = +6.5g - 4.2g (+6.5g is the gravitational force; pointing downward. -4.2g is from the tension of the string; pointing upward.)
F = (m1)a
+6.5g - 4.2g = 2.3g = (m1)a = 6.5a
a = 2.3g/6.5
v²= u² + 2aS
v² = 0² + 2(2.3g/6.5)(3.2)
with g = 10
v² = 2(2.3×10/6.5)(3.2) = 22.65
v = 4.76 m/s
with g = 9.8
v² = 2(2.3×9.8/6.5)(3.2) = 22.19
v = 4.71 m/s
2013-12-21 8:21 am
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佒儺凢
佒儺凢
收錄日期: 2021-04-27 20:49:14
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131219000051KK00154