✔ 最佳答案
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
