1.
If a 2000kg jet starts from rest and moves down a runway with its engine applying a force of 12000N and wind resistance applying a force of 2000N, how far will it be down the runway in 7 seconds ?
2.Let's say you're pulling on a garbage can with 100N of force and friction is opposing your movement with 50N of force. If you go from rest to 2.5m/s in 5 seconds, what is the mass of the garbage can ?
3.If you were on vacation in a Third World country, you might actually have to draw your drinking water from a well with a bucket. Assume the bucket and the water it holds has mass of 21 kg.
a. How much force would you have to supply to pull the bucket up at a constant speed?
b. If you were afraid of the locals and wanted to get the bucket up as fast as possible, how fast could you lift the bucket of water up the 30m well shaft if you were able to supply 300N of force ?
Newton second law
2012-02-29 4:08 pm
回答 (2)
2012-02-29 5:27 pm
✔ 最佳答案
1. By F=ma,12000 - 2000 = 2000a
a = 5ms^-2(downwards)
By s= ut+1/2at^2,
s = 0(7) + 1/2(5)(7)^2
s = 122.5m(downwards)
2.By v = u+ at,
2.5= 0 +a(5)
a = 0.5ms^-2
By F=ma,
100-50=m(0.5)
m=100kg
3a. By Newton's first law,any object with constant velocity has 0 net force.
Therefore,
magnitude of pulling force = magnitude of weight
= 21(10)
= 210N
You should apply 210N upwards in order to pull the bucket up at a constant velocity
3b. I am not sure for this question.....
Is the magnitude of friction same as that in 3(a)?
Just assume that they are equal
F=ma
300-210=21a
a=90/21ms^-2
By s= ut+1/2at^2
30= 0(t)+1/2(90/21)t^2
t=3.74s
參考: f4 science student
2012-02-29 5:15 pm
1) Net force on the jet = 10000 N
Hence acc. of the jet = 10000/2000 = 5 m/s^2
Applying s = ut + at^2/2 with u = 0, a = 5 and t = 7:
s = 5 x 7^2/2 = 122.5 m
2) Net force on the can = 50 N
Using v = u + at with v = 2.5, u = 0 and t = 5, we have a = 0.5 m/s^2
Hence the mass of the can is 50/0.5 = 100 kg
3a) Since the bucket is being pulled up at constant speed, net force on it is zero.
Hence force to be applied = weight of bucket = 21 x 9.8 = 205.8 N
b) Net force on the bucket = 300 - 205.8 = 94.2 N
Hence acc. of the bucket = 94.2/21 = 4.49 m/s^2
Using s = ut + at^2/2 with u = 0, s = 30:
30 = 4.49t^2/2
t^2 = 13.4
t = 3.66 s
Hence acc. of the jet = 10000/2000 = 5 m/s^2
Applying s = ut + at^2/2 with u = 0, a = 5 and t = 7:
s = 5 x 7^2/2 = 122.5 m
2) Net force on the can = 50 N
Using v = u + at with v = 2.5, u = 0 and t = 5, we have a = 0.5 m/s^2
Hence the mass of the can is 50/0.5 = 100 kg
3a) Since the bucket is being pulled up at constant speed, net force on it is zero.
Hence force to be applied = weight of bucket = 21 x 9.8 = 205.8 N
b) Net force on the bucket = 300 - 205.8 = 94.2 N
Hence acc. of the bucket = 94.2/21 = 4.49 m/s^2
Using s = ut + at^2/2 with u = 0, s = 30:
30 = 4.49t^2/2
t^2 = 13.4
t = 3.66 s
參考: 原創答案
收錄日期: 2021-04-20 12:15:51
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