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# Work done/power/energy question watch

1. Please could someone give me some guidance on this question:
The acceleration that a car of mass 1200kg can produce is limited by 2 factors, the frictional force between the driving wheels and road can't exceed 6000N and the engine has a maximum power output of 30kW. What is the shortest time in which it could reach a speed of 20m/m from rest?
Thanks in advance for any help

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2. (Original post by RebeccaLM)
Please could someone give me some guidance on this question:
The acceleration that a car of mass 1200kg can produce is limited by 2 factors, the frictional force between the driving wheels and road can't exceed 6000N and the engine has a maximum power output of 30kW. What is the shortest time in which it could reach a speed of 20m/m from rest?
Thanks in advance for any help
You should know the relevant relationships between work, force, and distance and power, force, and velocity i.e. that:

It should be clear that we can minimise the time taken to reach 20 m/s by maximising the acceleration at all times.

However, if we try to rev the engine up to 30 kW when the car is only going at 1 m/s, say, we'll find that the force exerted on the road is . This is greater than the maximum 6000N frictional force that the road-tyre surfaces can generate, so the wheels will spin.

So for lower speeds, there is a maximum power output from the engine that we can use: it's the output that ensures that the frictional forrce doesn't exceed 6000N.

At higher speeds, say, 10 m/s then a similar calculation gives , so we can safely drive the engine at full tilt, giving us maximum acceleration without spinning the wheels.

So, some questions:

1. What is the speed at which we can drive the engine at 30 kW without spinning the wheels?

2. Above this speed, what force will the car feel, as a function of ?

3. What is the maximum force we can apply to the car in the low speed regime?
3. Okay so you can drive the engine at 5m/s or greater when the engine power is 30kW, the function of the force after that is 30,000/v and the force before that is 6000N.....is that all right?

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4. (Original post by RebeccaLM)
Okay so you can drive the engine at 5m/s or greater when the engine power is 30kW, the function of the force after that is 30,000/v and the force before that is 6000N.....is that all right?

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Yes, that looks fine (though I wouldn't use your wording "drive the engine at 5 m/s" - I'd put it the other way round: "drive the engine at 30 kW when the speed is 5 m/s or greater").

I haven't completed the question but it looks like you'll have a simple DE to solve. What board is this, BTW?
5. Yeah sorry, that wording wasn't great! And let me know when you get an answer it's OCR

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6. (Original post by RebeccaLM)
And let me know when you get an answer
It'd be more useful to you if you found the answer, and I checked it, I think.
7. I know the answer from the textbook but I'm still struggling with how to go about it :/ so far I'm thinking that if the maximum force from 0 to 5m/s is 6000N then that makes the acceleration for that time period constant so it takes 1 second to accelerate from 0 to 5m/s. Then after that I tried to use work done by forces = gain in kinetic energy. So 30000t = 0.5x1200x(15^2) which gives 30000t=135000 so t is 4.5, so overall t is 5.5..... But this isn't the answer in the textbook

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8. Is it 16 seconds?
9. The answer in the book says (I'll spoiler it just in case people want to work through it themselves)
Spoiler:
Show

8.5 seconds

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10. Finally worked it out guys

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11. (Original post by RebeccaLM)
Finally worked it out guys

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?
12. (Original post by atsruser)
?
So it takes 1 second to accelerate from 0 to 5m/s because the maximum friction is 6000N and the mass of the car is 1200kg so F=ma gives a=5ms^-2..... Then I used work done = gain in KE so 30000t = 1/2 x 1200 x (20^2-5^2) so t for that is 7.5 therefore the total time is 8.5

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13. (Original post by RebeccaLM)
So it takes 1 second to accelerate from 0 to 5m/s because the maximum friction is 6000N and the mass of the car is 1200kg so F=ma gives a=5ms^-2..... Then I used work done = gain in KE so 30000t = 1/2 x 1200 x (20^2-5^2) so t for that is 7.5 therefore the total time is 8.5

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Right. Well done. You noticed something that I didn't initially - you can use an energy argument rather than solve a DE, though the DE would have come to the same thing in the end:

which is what you started with.

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