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high school physics student help |
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| Aug19-07, 02:06 AM | #1 |
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high school physics student help
i am currently momentum, work, energy and power in physics and i am currently confused in a few concepts.
This question states that: A car with mass 1500kg accelerates from 0 to 100km/hr in 4 seconds. calculate the power output of the car. -- My teacher states to solve this question using kinetic Energy = Work , and equating this into the power formula... is there a reason why you have to use this approach to solve the question??? when i try and figure it out , i use momentum and impact and i get a completely different answer. The thing i struggle with is if i get given a question, which method do i use to solve it?? thankyou |
| Aug19-07, 02:52 AM | #2 |
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There are a number of different ways to solve any problem, but in classical mechanics energy/work is often the best (easiest) way.
Can you show your calculations for using a momentum approach? |
| Aug19-07, 04:09 AM | #3 |
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values:
m=1500kg a=25ms-2 v=27.78ms-1 t=4 P=mv =1500 x 27.78 = 41670 SN Impulse = (delta) P f x 4 = 41670 F = 10417.5 N W = FS = 10417.5 x 111.12 = 1157592.16 J P = w/t = 1157592.16/4 = 289398.15 Watts the correct answer is: 144675.9 Watts (found using {delta}E = Work) |
| Aug19-07, 04:37 AM | #4 |
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high school physics student help
You converted the velocity from km.h-1 to m.s-1 which is a good idea.
The acceleation doesn't equal 25 ms-2. It does equal 25 km.h-1.s-1. But it looks like you didn't use that value later on, so that mistake didn't matter ![]() Your value of S looks wrong if it's the total distance travelled. The average velocity is (0+27.78)/2 = 13.89 ms-1. The distance travelled in 4s = 4*13.89 m. There is nothing in the original question that says the acceleration is constant, and the answer doesn't depend on it being constant. So the energy method is better (as well as being quicker) because you are not making an unjustified assumption. But since constant acceleration is one valid way of meeting the conditions of the questions, using it will give you the right answer. To see that constant acceleration isn't necessary, try assuming the car does 0-100 in 2 sec and then travels at constant speed for 2 sec. The AVERAGE power will come out the same. The question should really have said "find the average power output", not "find the power output" because the power varies with time (try it - calculate the power for the first 2s and the last 2s, using your constant acceleration assumption). |
| Aug19-07, 05:57 AM | #5 |
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yes i've realized my error... he accelerates in 4 seconds to that speed, and i was using the D=ST formula which is incorrect!.. as his speed isnt constant for a period of time...
now i understand why the energy method is more accurate.. |
| Aug19-07, 06:04 AM | #6 |
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ok ive tried the approach with the constant acceleration... this time i used the average velocity example of 18.89 ms-1 .. then i calculated the displacement using the s=ut+1/2at2 .. then i worked out the work and the power came to 133816.76 Watts... very close to the final answer!
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| Aug21-07, 04:17 AM | #7 |
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Ek = (0.5)mv^2
= (0.5) . 1500 . 27.78 . 27.78 = 578,796.3 P = W/t = 578,796.3 / 4 = 144,699.075 W :) |
| Aug21-07, 04:28 AM | #8 |
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p = mv
1500 . 27.78 = 41,670 kgms^-1 dp/dt = F 41,670/4 = 10,417.3N W=Fd = 10,417.3 . d d = Vi.t + (0.5)a.t.t d = (0.5).(27.78 / 4).4.4 d = 55.56 W = Fd = 10,417.3 . 55.56 = 578,785.188 P = W/t = 578,785.188 / 4 = 144,696.297 W |
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