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Anonymous users2024-02-07The mass of the car is m = 2000kg, the rated power p = 80kw, the maximum speed is vm = 20m s, so that the car starts from a standstill on the highway with an acceleration of 2m s2 to do a uniform linear motion, if the resistance of the car is constant, find (g = 10).
1.The resistance to which the car is subjected.
f=p/v=80000/20=4000n
2.How long can this uniform acceleration process last.
3.The instantaneous power of the car's engine at the end of 3s after the start of movement.
v=at=2*3=6m/s
p=fv=8000*6=48000w=48kw4.What is the acceleration of a car when the velocity of the car is?
f=p/v=80000/
a=(f-f)/m=(6400-4000)/2000=5.How fast is the car when the acceleration of the car is.
f=ma+f=2000*
v=p/f=8000/5000=16m/s
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Anonymous users2024-02-06Resistance f=, rated output power p=80kw
1) When the car is driving at the maximum driving speed vm, the traction force f is balanced with the resistance f, p = fvm, vm = 40m s
2) The speed at the end of the start-up phase of the car is V
When the car starts at a uniform acceleration, the traction force of the car is f'，f'-f=maf'=f+ma=4*10^3n
At the end of the uniform acceleration start stage, the power of the destruction is the rated output power of p=80kwp=fv, v=20m s
3) When the speed is V1=5M s, the instantaneous power of the traction force of the car is P5M S<20M S, and the traction force is F in the process of uniform acceleration at this time'
p=f'v1=2*10^4w
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Anonymous users2024-02-05Analysis: (1) No, when the car moves on an inclined plane, the force of the oblique downward is:
F oblique = f + GSIN =, so it can start to keep the traction force of 8000N unchanged on the slope, when the power reaches the quota power of 60kw, the speed increases, and the traction force will decrease.
2) When the traction force is equal to the force sliding down the inclined plane, the speed reaches the maximum, and the maximum speed is:
VM=P F oblique = 60000W 6000N = 10M S (3) In the process of going uphill with the traction force of 4000N, the car does a uniform deceleration movement, and the speed is the largest and the power is the largest at the beginning
PM = FVM = 4000 * 12W = 48000W The average power is: P average = FV average = 4000N * (12M S + 4M S ) 2 = 3200W
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Anonymous users2024-02-04Analysis: Whether the car can maintain traction at 8000 n uphill There are two things to consider: first, is the traction greater than the drag?
Second, if the car keeps accelerating, whether its power will exceed the rated power, according to p=fv solution. This question examines the calculation of the constant power of a car's traction. Many students immediately came to the conclusion after getting F> F + mgsin:
The car can keep the traction force 8000 n uphill constant; Without taking into account the fact that the speed of the car is constantly increasing due to acceleration, its power is constantly increasing, and if the slope is long enough, this mode of movement is not allowed. Solution: Analyze the force of the car during the uphill process as shown in the figure
Traction force f, gravity mg 4 104 N, f kmg 4 103 N, support force n, according to the title sin 5 100. (1) when the car goes uphill, if F 8000N, and F MGSIN 4 103 4 104 1 20 6 103 N, i.e. F> F + MGSIN, the car will accelerate uphill, the speed continues to increase, and its output power P=FV also continues to increase, after a long time, will exceed its rated output power, so, the car can not keep the traction force for 8000N unchanged uphill. (2) when the car goes uphill, the speed is getting bigger and bigger, and the traction force must be continuously reduced to ensure that the output power does not exceed the rated output power, when the traction force f = f + mgsin = 6 103 n, the car acceleration is zero, the speed increases to the maximum, and is set to vm, then p fv (f mgsin)·vm; f= f + mgsin =6 103 n (3) If the traction force f=4000N, when the car goes uphill, the speed is constantly decreasing, so the initial power is the maximum, P=fv=4000 12=48 103W.
The average power over the entire process is =32 103W
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Anonymous users2024-02-03(1) The resistance f=
vmax p f=60 10 3 3920=.
2）a=x=1/2at^2=1/
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Anonymous users2024-02-02Hello The solution is as follows:
1. The driving resistance of the car f=m*g*k=
Then vmax=po f=60x10 3 4000=15m s2, root question: a=vmax t=15
then x=1 2xaxt 2=1
Hope it helps
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Anonymous users2024-02-01(1) When the car has the maximum speed, the traction force and the resistance are balanced at this time, which can be obtained:
p=fv=fvm, so vm pfpkmg
m/s=20m/s；
2) When the velocity v=10m s, then f=p
v 4 110n 4000n, according to Newton's second law, f-f=ma, solution a=f?fm＝4000?2000
1m s (3) If the car starts from a standstill to do uniform acceleration linear motion, when the rated power is reached, the uniform acceleration phase ends
From f-f=ma, the traction force f1 = f + ma1 = 2000 + 2000 1n = 4000n is solved by f-f=ma, and the final velocity v pf of the uniform acceleration movement is solved
10m/s；
Then the time of the acceleration motion t v
a＝101＝10s．
Answer: (1) The maximum speed vm that the car can achieve on the road surface is 20m s (2) The acceleration a when the speed of the car is 10m s is 1m s2 (3) The time that this process can last is 10s
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Anonymous users2024-01-31The engine power p=30kw, and it can move forward at a constant speed at the maximum slag loss rate of v=15m s on the horizontal road surface like Doushen.
f=p/vm=2000n
When the rate is reduced to v'=10m s.
f=p/v=3000n
The acceleration of the car is a=f-f=ma a=
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Anonymous users2024-01-301) Let the traction force be f, f-f=f-2000=ma=2000, so f=4000 and f is constant at 4000 because of uniform acceleration
In the uniform acceleration stage, when the engine power reaches the rated 40kw, the uniform acceleration phase ends.
So the maximum speed v = p divided by f = 40kw 4000 = 10m st = v a = 10 s
2) Then the car drives at constant power, because p=fv, p is unchanged, to make v bigger, only let f decrease, the car is accelerated to change, the acceleration will be smaller and smaller, when f is small to =f, there is a maximum speed vm, at this time, f=2000=f
So vm = p 2000 = 40kw 2000 = 20m s3) after that the car does a uniform motion, the maximum speed is 20m s
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Anonymous users2024-01-29Solution: Initial to uniform velocity, by the kinetic energy theorem.
pt-fs = open mask.
When evenly blocking the alarm speed simple letter cover, p=fv
Newton's third law.
f=f gives s= 1292m
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