Energy and Work Energy and Work Physics Help Forum 
Jun 8th 2015, 01:46 PM

#1  Junior Member
Join Date: Oct 2010
Posts: 21
 The power of a drive motor
We have a car with a total mass = 1500 kg.
We need to maintain a constant speed of 90km/h when driving on a hill with a slope of 10%.
How much power in kW the car needs to have?
So I have converted the 90km/h to 25m/s, the slope is 45° i think. We don't know the accelaration. Do we need it? For, f = ma...
How to continue? Thank you for your help.

 
Jun 8th 2015, 02:08 PM

#2  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 862

the slope is 45° i think.

Think again about this first

 
Jun 9th 2015, 10:19 AM

#3  Junior Member
Join Date: Oct 2010
Posts: 21

Oh, sorry typo: I meant 4.5°. If the full slope is 45°, if it's not and is logically 90°, then 10% is 9°. But this is not the main problem here.
How do I get the Force here? F = ma, we know the mass, but not the acceleration. And how much distance did the car move. With that we can get kinetic energy: force*distance*cos(4.5)?
Thanks for helping me.

 
Jun 9th 2015, 10:55 AM

#4  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 862

Well my calculator makes it 5.7 degrees.
What do you think a gradient of 10% means?
You still need to get this right before you can work anything else out.
Then you can look at the right angled triangle, formed by the change in vertical distance, the change in horizontal distance and the actual distance travelled. http://en.wikipedia.org/wiki/Grade_(slope)

 
Jun 9th 2015, 11:11 AM

#5  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,280

Originally Posted by Nforce How do I get the Force here? F = ma, we know the mass, but not the acceleration. 
You said the speed is constant, so a = 0. The forces you need to overcome are gravity (due to the 10% slope), plus drag from wind resistance (which depends on the car's geometry and speed) plus friction forces  are you supposed to ignore those effects? At 90 KPH the wind resistance will be quite substantial.
Originally Posted by Nforce And how much distance did the car move. With that we can get kinetic energy: force*distance*cos(4.5)? 
No. Force times distance = work done, but you asked about power, not work. Power is calculated from Force x Velocity. You know the velocity, but you still need to determine what that force is, which is caused by the weight of the car operating on a 10% grade plus the drag and friction effects I mentioned earlier.
Last edited by ChipB; Jun 9th 2015 at 02:18 PM.

 
Jun 10th 2015, 07:55 AM

#6  Junior Member
Join Date: Oct 2010
Posts: 21

OK, thank you for the relation between speed and force.
Now check my calculation.
First the slope of the hill is 5.7°, we compute that with tan(0.1/1) ~ 5.7
We ignore the friction and wind resistance. It's a high school problem, so I think they ignored that.
F = m*g*v*cos(5.7)
So:
1500kg*9.8*25*cos(5.7) is approx 37 kW. Is my calculation correct?
Thank you for your help.

 
Jun 10th 2015, 09:40 AM

#7  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,280

Originally Posted by Nforce F = m*g*v*cos(5.7)
So:
1500kg*9.8*25*cos(5.7) is approx 37 kW. Is my calculation correct? 
Not quite correct. Why did you decide to use cosine, instead of sine? Think about it  if the slope was steeper, you would expect the force to be greater (it's harder to push a car up a steep hill than on a level road). But the cosine function is max at an angle of 0 degrees and goes to zero at 90 degrees. If you get confused on whether to use sine or cosine it can be helpful to think about the extremes of slope at 0 degrees or 90 degrees  if you use cosine you would find that it takes zero force to push something up a vertical wall, which is clearly not right.
Also, you said you calculated about 37 kW, but your calculation using cosine actually gives more like 367 kW  watch those powers of ten!
Last edited by ChipB; Jun 10th 2015 at 02:29 PM.

 
Jun 10th 2015, 12:18 PM

#8  Junior Member
Join Date: Oct 2010
Posts: 21

You are right. But with sin I get 202 kW. Why do I get minus?

 
Jun 10th 2015, 12:27 PM

#9  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,280

Originally Posted by Nforce You are right. But with sin I get 202 kW. Why do I get minus? 
It seems you calculated sin(5.7), where 5.7 is in units of radians, not degrees. You can convert 5.7 degrees to radians by using 5.7 x pi/180.

 
Jun 10th 2015, 12:33 PM

#10  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 862

Are you just using chipB's formula or do you understand what is going on?
I said you could do this with a particular right angled triangle.

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