Car going up the slope, Power of Engine

In summary, the problem is asking for the power of a car engine going up a slope with a constant acceleration and a given friction force. The power is calculated by dividing the work done by the time taken, and the work is the sum of the forces (gravity, friction, and acceleration) multiplied by the distance traveled. The power requirement is a function of velocity, mass, acceleration, and the angle of the slope, and increases as any of those values increase.
  • #1
CroSinus
32
1

Homework Statement


Hello everybody,

can you help me with the following problem please. A car is going up the slope with a constant acceleration a (unknown). The friction force equals to Ffr = const. + 0,7v^2, where v is the velocity of the car. The angle of the slope is given and equals [itex]\theta[/itex]. The mass of the car is given and equals m. What is the power of the car engine?

Homework Equations


The Attempt at a Solution


Homework Statement


Homework Equations


The Attempt at a Solution

 

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  • #2
You must first show your attempt before anyone on this forum can help. That said, this is a problem where keeping aware of units behooves you.
 
  • #3
Well, here's my attempt. The car engine has to overcome three forces. The first one is the force of the slope which equals G*sin theta. The second one is the friction force which is velocity dependent and equals Ffr = const. + 0,7v^2. The car engine has also to accelerate the vehicle. So the accelerating force equals to m*a. Where m is the car mass and a is the acceleration of the car.

In order to find the power of the engine I write: P = W/t. Power equals work divided by time. So I have three forces: G*sin theta + const. + 0,7*v^2 + m*a and I need the path of the car calculated as the length of the slope on which the car is going upwards.

Am I on the right path? Any hints please?

Thanks,
CroSinus
 
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  • #4
Hint:

In SAE units, power is ft-lbs/sec which can also be written as lbs-(ft/sec).
 
  • #5
Thanks for trying to help me.

Well, power is defined as P = work done / (time taken to do work). P = dW/dt.
dW = F*dr
P = F dr/dt = F*v

Can you explain to me the meaning of the equation P = F*v. I think I can use this equation only when the Force is constant. But in my case F is changing with speed (F = cons. + 0,7v^2), or better to say with time. How can I apply the equation P = F*v then? Please help me!

Thank you very much,
CroSinus
 
  • #6
Your power requirement will have several components. One is drag, another is the fact that it is accelerating, another is due to the vehicle increasing its potential energy.

Because you have drag as a function of velocity, the power requirement necessarily must be a function of velocity. It must increase as speed increases due to friction force increasing with velocity. A similar argument can be made for the rate of change of potential energy which is power.

"Can you explain to me the meaning of the equation P = F*v. I think I can use this equation only when the Force is constant."

If you stick your hand out of a car window when the car is moving 60 mph (88ft/sec), the power used is 88*F ft-lbs/sec. F would come from drag coefficient, air density, frontal area, and the velocity squared. So the force is speed dependent which makes the power requirement a function of the velocity cubed.
 
  • #7
So the force is speed dependent which makes the power requirement a [U said:
function of the velocity cubed[/U].

Thank you very much for you answer. But how can I calculate power from the data given in my example. A) angle of the slope (theta) B) Ffr = 218 N + 0,7v^2 C) car mass = m

Thanks for help,
CroSinus
 
  • #8
Because they do not provide you with actual numbers, your answer will have to be in the form of a function of V,m,theta, and a. Obviously, if theta or mass or acceleration increase, the power requirement will increase.

If theta is larger, the rate of change of potential energy increases.
If acceleration is greater, the power requirement to cause that increases due to Newton's second law.
If the vehicle has greater mass, the power requirement increases due to rate of change of potential energy as well as Newton's second law (more mass involved).
As V increases, the power requirement increases due to drag and rate of change of potential energy.

If they provided you with the actual numbers for theta, V, mass, and acc., you could come up with a number for the power needed at that instant in time.
 
  • #9
Great. Thank you very much again.
 

FAQ: Car going up the slope, Power of Engine

How does the power of the engine affect a car going up a slope?

The power of the engine plays a crucial role in how a car performs on a slope. A higher power engine will have more horsepower and torque, allowing it to exert more force and overcome the resistance of the slope more easily. On the other hand, a lower power engine may struggle to maintain speed and may even stall on a steep slope.

What factors contribute to the power of an engine?

The power of an engine is determined by several factors, including its size, design, and fuel efficiency. Generally, a larger engine will produce more power due to its ability to burn more fuel and air. The design of the engine, such as the number of cylinders and the use of turbochargers, can also impact its power output. Additionally, a more fuel-efficient engine will have a higher power-to-weight ratio, allowing it to generate more power with less fuel.

Can a car with a smaller engine still perform well on a slope?

Yes, a car with a smaller engine can still perform well on a slope, depending on the slope's steepness and the car's weight. A smaller engine may struggle on a steep slope, but it can still maintain a steady speed on a gradual incline. Additionally, a lighter car with a smaller engine may be able to overcome the resistance of a slope more easily than a heavier car with a larger engine.

How does the power of an electric car's motor compare to a traditional gasoline engine?

An electric car's motor produces power differently than a traditional gasoline engine. While a gasoline engine relies on combustion to generate power, an electric motor uses electrical energy stored in the battery. In general, electric motors have a higher power output than similarly sized gasoline engines, but they may have limitations in maintaining power for extended periods.

Does the power of the engine impact a car's fuel efficiency?

Yes, the power of the engine can impact a car's fuel efficiency. Generally, a more powerful engine will consume more fuel, resulting in lower fuel efficiency. However, advancements in engine technology and design have allowed for more powerful and efficient engines, meaning that a car with a higher power engine may still have good fuel efficiency compared to older models.

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