質問: s.4 phy 1. a monkey falls from rest from a tall tree. how far does it fall during the third second? assume air resistance is negligible. answer is 25m. why? 2. a car keeps slowing down uniformlyalong a straight road. it slows down to 20 ms^(-1) after travelling a distance of 100 m, and runs 80 m further before it comes to a stop. what is its speed at first? answer is 30 ms^(-1). why? 3. cn tower in canada is 533 m tall and it is the tallest building in the world. if a stone of 200 g is thrown vertically upwards at 5 ms^(-1) at the top of the cn tower, what is the kinetic energy of the stone when it drops to the groung? neglect air resistance. answer is 1069j. why?
回答: 3. when the small stone was thrown. it had a k.e. of (1/2)(0.2)(52). it goes up and change this k.e. to p.e. when the small stone comes down to a height of 533 m. the p.e. gained returns back to k.e. the total energy of the small stone is (1/2)(0.2)(52) + mgh = (1/2)(0.2)(52) + (0.2)(10)(533) = 1068.5 j 2014-03-09 17:05:37 補充: 1. a monkey falls from rest from a tall tree. how far does it fall during the third second? assume air resistance is negligible. answer is 25m. why? in the first 2 seconds, it falls : h = v0t + (1/2)(g)t2 = (0)(2) + (1/2)(10)(2)2 = 20 m 2014-03-09 17:07:02 補充: in the first 3 seconds, it falls : h = v0t + (1/2)(g)t2 = (0)(3) + (1/2)(10)(3)2 = 45 m it falls (45 - 20) = 25 m in the third second 2014-03-09 17:11:08 補充: 2. a car keeps slowing down uniformlyalong a straight road. it slows down to 20 ms^(-1) after travelling a distance of 100 m, and runs 80 m further before it comes to a stop. what is its speed at first? answer is 30 ms^(-1). why? use (vf)2 = (vi)2 - 2as 202 = (vi)2 - 2a(100) ... (1) 2014-03-09 17:15:25 補充: it comes to a stop means v=0 sub into the formula again 02 = 202 - 2a(80) a = 2.5 m/s2 put (a = 2.5 m/s2) into (1) ==> 202 = (vi)2 - 2(2.5)(100) vi = 30 m/s 2014-03-13 20:30:01 補充: 1. 猴子從高樹跌下,在第 3 秒下跌了多小米。 0 至 1 秒是 第 1 秒 1 至 2 秒是 第 2 秒 2 至 3 秒是 第 3 秒 在頭 2 秒 下跌了 h = v0t + (1/2)(g)t2 = (0)(2) + (1/2)(10)(2)2 = 20 m 在頭 3 秒 下跌了 h = v0t + (1/2)(g)t2 = (0)(3) + (1/2)(10)(3)2 = 45 m 在 第 3 秒 下跌了 (45 - 20) = 25 m 2014-03-13 20:36:24 補充: 2. 小車在直路上持續减速,它在行駛 100 米後減速至 20 米/秒,再前進 30 米之後完全停定,求小車初速。 公式 : (vf)2 = (vi)2 - 2as 其中 : vf = 末速 vi = 初速 a = 加速度 s = 位移 (前進距離) 2014-03-13 20:39:19 補充: 用 行駛 100 米後減速至 20 米/秒 代入 202 = (vi)2 - 2a(100) ... (1) 用 再前進 80 米之後完全停定 代入 停定 即 vf = 0 02 = 202 - 2a(80) a = 2.5 m/s2 用 (a = 2.5 m/s2) 代入 (1) ==> 202 = (vi)2 - 2(2.5)(100) vi = 30 m/s