Physics Forces!

2013-09-30 5:42 am
Q1:
The graph for Q1:
http://blog.yimg.com/3/KcutVhp7s58pQ5Vx_PowSzy3i98NvOevXdlckjayvtQThTvtxzASgQ--/83/l/1L.QhxT.cOjX7U44Pc6A0Q.jpg
A 1.90kg object initially at rest at the origin is subjected to the time-varying force shown below. What is the object's velocity at t total=12.0s if the maximum force applied in the x-component, Fx, is 18.0N?


Q2:
The graph for Q2:
http://blog.yimg.com/3/KcutVhp7s58pQ5Vx_PowSzy3i98NvOevXdlckjayvtQThTvtxzASgQ--/84/l/4zNGdBbXjqFcFlILNwHh2g.jpg
Henry gets into an elevator on the 50th floor of a building and it begins moving at t=0.00 s. His apparent weight is shown over the next 36.0s.
a)What is Henry's mass if his maximum apparent weight within the elevator is 700.0N?
b)How far has Henry traveled after 36.0s?

Can anyone solve the above question?
Every time I saw this kind of graph, I don't have any idea how to solve the question, is there any techniques to read the graph?

回答 (2)

2013-09-30 8:03 pm
✔ 最佳答案
Q1. The force is constant, equals to 18 N, from t = 0 to 8 s
Hence, acceleration = 18/1.9 m/s^2 = 9.474 m/s^2
Velocity at t = 8 s is 9.474 x 8 m/s = 75.79 m/s

The force decreases linearly from t = 8 to 12 s,
hence, the mean force = 18/2 N = 9 N
Mean acceleration = 9/1.9 m/s^2
Increase in velocity = (9/1.9) x (12-8) m/s = 18.95 m/s
Velcoity at t = 12 s is (75.79 + 18.95) m/s = 94.74 m/s

Q2. (a) Based on the given graph, the elevator is descending. The max apparent occurs when the elevator decelerates and comes to a stop, which is indicated between the time t = 30 to 36 s.

Thus from t = 6 to 30 s, the elevator is descending at constant speed. During such time, the apparent weight equals to the real weight.
Hence, the real weight = 700 x (3/4) N = 525 N
Mass of Henry = 525/g kg = 52.5 kg (take g = 10 m/s^2)

(b) Apparent weight from t=0 to 6 s is 700 x (2/4) N = 350 N
Acceleration = (525 - 350)/52.5 m/s^2 = 3.33 m/s^2
Distance travelled = (1/2).(3.33).(6^2) m [ use: s = ut + (1/2)at^2]
= 60 m
Speed at t = 6 s is 3.33 x 6 m/s = 20 m/s

From t = 6 to 30 s,
Distance travelled = 20 x (30-6) m = 480 m

From t = 30 to 36 s,
Deceleration = (700 - 525)/52.5 m/s^2 = 3.33 m/s^2
Distance tracelled = 20 x 6 + (1/2).(-3.33).(6^2) [ s = ut + (1/2)at^2 ]
= 60 m

Therefore, total distance travelled
= (60 + 480 + 60) m
= 600 m






2013-09-30 6:08 am
從所有你可以聯想到的東西思考,當你想到 F = ma,這個圖便相當於一個 加速度 a 對時間 t 的圖,而面積便是速度 v 了。Fmax = 18 N,a(max) = 18 / 1.9 = 9.47 m / s^2。面積 = (上底 + 下底)X高 / 2 = ( 8 + 12 ) X 9.47 / 2 = 94.7 米。

2013-09-29 22:13:21 補充:
其實如果從未見過都要識做是要有很深的功夫才成的,多做習題,多見識各種題目類型當然會很有幫助。

Sorry ,上面的單位錯了,應該 是 94.7 m / s 才對。

2013-09-29 22:39:03 補充:
我對 第 2 題 這樣想 F = ma,縱座標軸(垂直) 分為 4 個刻度,相當於 700 N,每個刻度為 700 / 4 = 175 N。橫座標每個刻度為 6 秒,要假設 0 秒到 6 秒, a = 0,(這個假設似乎免強,但若不如此假設便無從得知 Henry 的 mass 了。Henry 的 mass 做成 350 N 的重力,所以 m = 700 / g = 700 / 9.8 = 71.4 kg。

2013-09-29 22:54:42 補充:
從 6 秒至 30 秒為一個等加速度運動 constant acceleration motion
F = 525 - 350 = 175 N
175 = (71.4)a
a = 2.45 m/s^2
S = Vo t + 0.5at^2 = 0 + 0.5(2.45)(30-6)^2 = 706 m

第 30 秒時的速度 =
V(30) = 0 + at = 0 + (2.45)(24) = 58.8 m/s

2013-09-29 22:58:50 補充:
第 30 至 36 秒,用以加速的力為 350 N,a = 4.9 m / s^2
S = Vo t + 0.5at^2 = 58.8 X 6 + 0.5(4.9)(6)^2 = 441 m

總位移 = 441 + 706 = 1147 m

1147 / 3 = 382 層樓,再加上在 50 樓入電梯,邊有大廈咁高架?


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